NEET Physics: Electromagnetic Waves and Optics · Chapter 2

Ray Optics and Optical Instruments

What to remember

Cartesian sign convention: all distances are measured from the pole of the mirror or the optical centre of the lens. The direction of the incident light is taken as positive. Heights above the principal axis are positive and heights below are negative.
Mirror and lens formulae: mirror 1/v + 1/u = 1/f with m = −v/u; thin lens 1/v − 1/u = 1/f with m = v/u. Always substitute u, v, f with their signs.
Refraction and total internal reflection: n = sin i / sin r. Total internal reflection needs light to go from a denser to a rarer medium at an angle of incidence larger than the critical angle, where sin ic = 1/n.
Prism and instruments: n = sin((A + δm)/2) / sin(A/2); thin prism δ = (n − 1)A. A telescope in normal adjustment has m = fo/fe and length fo + fe. A compound microscope has m = (L/fo)(D/fe) for the final image at infinity.

Reflection by spherical mirrors

A concave mirror has its reflecting surface curved inward and it converges a parallel beam. A convex mirror curves outward and diverges a parallel beam. The principal focus F is the point where rays parallel to the principal axis meet (concave) or appear to come from (convex). The focal length f is the distance PF, and for a small aperture

f = R/2

where R is the radius of curvature.

Sign convention (Cartesian).

The pole P is the origin. The principal axis is the x-axis. The incident light travels along +x.
Distances measured in the direction of incident light are positive. Distances against it are negative.
For a real object in front of a mirror, u is negative.
A real image in front of a mirror has v negative. A virtual image behind a mirror has v positive.
For a concave mirror, f and R are negative. For a convex mirror, f and R are positive.
Heights measured upward from the axis are positive; downward are negative.

Mirror equation and magnification

1/v + 1/u = 1/f

m = h′/h = −v/u

A negative m means the image is real and inverted. A positive m means the image is virtual and erect. |m| greater than 1 means enlarged.

Mirror and object position	Image position	Nature and size
Concave, object beyond C	Between F and C	Real, inverted, diminished
Concave, object at C	At C	Real, inverted, same size
Concave, object between C and F	Beyond C	Real, inverted, enlarged
Concave, object at F	At infinity	Real, inverted, highly enlarged
Concave, object between F and P	Behind the mirror	Virtual, erect, enlarged
Convex, any position of real object	Between P and F, behind the mirror	Virtual, erect, diminished

For a lens (see below): a convex lens forms a real, inverted image when the object is beyond its focus and a virtual, erect, enlarged image when the object is within the focal length. A concave lens always forms a virtual, erect, diminished image of a real object.

A convex mirror is used as a rear-view mirror because it gives an erect, diminished image with a wide field of view. A concave mirror is used in shaving mirrors and in torch reflectors.

Refraction of light

When light passes from one medium to another, it changes direction at the boundary. This is refraction. Snell's law: the incident ray, the refracted ray and the normal lie in one plane, and

sin i / sin r = n₂₁ = n₂/n₁

The refractive index of a medium with respect to vacuum is n = c/v, where v is the speed of light in the medium. When light enters a denser medium (larger n), it bends toward the normal and its speed and wavelength decrease, but its frequency stays the same.

Approximate refractive indices: water 1.33, crown glass 1.5, diamond 2.42.

Real and apparent depth. An object in a denser medium of index n, seen from the rarer medium (for example air) with the eye nearly directly above it, appears at

apparent depth = real depth / n

So a coin under water looks closer to the surface than it is.

Lateral shift in brief. A ray passing through a parallel-sided glass slab emerges parallel to its original direction but is shifted sideways. For a slab of thickness t, the shift is t sin(i − r)/cos r. For near-normal viewing, an object seen through a slab of index n appears raised by t(1 − 1/n).

Total internal reflection

When light goes from a denser medium to a rarer medium, the refracted ray bends away from the normal. At the critical angle ic the refracted ray grazes the boundary (angle of refraction 90°). For ic from a medium of index n into air, and between two media with n₁ greater than n₂:

sin ic = 1/n (into air) and sin ic = n₂/n₁ (between media)

If the angle of incidence is larger than ic, there is no refracted ray and all light is reflected back. This is total internal reflection (TIR). Two conditions are needed: light must travel from denser to rarer medium, and i must be greater than ic.

Medium (to air)	n	Critical angle (approx.)
Water	1.33	49°
Crown glass	1.5	42°
Diamond	2.42	24°

Applications of TIR:

Sparkling of diamond: its critical angle is small, so most light entering undergoes repeated internal reflections before leaving, and the stone flashes brightly.
Optical fibres: a glass core of higher index is surrounded by a cladding of lower index. Light is guided along the core by repeated TIR with very little loss. They are used in communication and in endoscopy.
Prisms used to bend light: a right-angled isosceles prism turns a ray through 90° (the ray strikes the long face at 45°, more than ic of glass) or through 180°, with no loss as in a mirror.
Mirage: on a hot day the air near the ground is hotter and less dense than the air above, so light from the sky bends upward and totally reflects, and a distant object appears inverted as if reflected in water.

Refraction at spherical surfaces and thin lenses

For a single spherical refracting surface separating medium 1 (index n₁, object side) from medium 2 (index n₂):

n₂/v − n₁/u = (n₂ − n₁)/R

R is positive if the centre of curvature lies on the side of the outgoing light.

Thin lens formula and magnification

1/v − 1/u = 1/f

m = h′/h = v/u

A convex (converging) lens has positive f. A concave (diverging) lens has negative f. For a real object, u is negative. A positive m means an erect, virtual image. A negative m means an inverted, real image.

Lens maker's formula (lens of index n in air, R₁ for the first surface, R₂ for the second):

1/f = (n − 1)(1/R₁ − 1/R₂)

For an equiconvex lens R₁ = +R and R₂ = −R. For a plano-convex lens with the curved face toward the incident light, R₁ = +R and R₂ = infinity.

Power of a lens: P = 1/f, with f in metre. The unit is the dioptre (D). A convex lens has positive power and a concave lens has negative power. A lens of focal length 50 cm has power +2 D.

Lenses in contact (thin lenses placed together):

1/f = 1/f₁ + 1/f₂ and P = P₁ + P₂

The total magnification is m = m₁ × m₂.

Refraction through a prism and dispersion

In a prism of angle A, a ray enters at angle of incidence i, refracts to angle r₁, meets the second face at r₂, and leaves at angle e. The relations are

r₁ + r₂ = A and δ = i + e − A

The deviation δ first decreases and then increases as i increases. At the minimum deviation δm, i = e and r₁ = r₂ = A/2. In this position the ray inside an isosceles prism runs parallel to the base. Then

n = sin((A + δm)/2) / sin(A/2)

For a thin prism (small A and small angles), the deviation is almost independent of i:

δ = (n − 1)A

Dispersion. The refractive index of glass depends on wavelength: it is larger for violet and smaller for red. A prism therefore bends violet the most and red the least, and white light splits into its colours. In vacuum all colours travel at the same speed, but in glass red travels faster than violet.

Optical instruments

Take the least distance of distinct vision (near point) as D = 25 cm.

Simple microscope (a single convex lens of small focal length f used as a magnifier):

Image at the near point: m = 1 + D/f.
Image at infinity: m = D/f.

Compound microscope. It has two convex lenses. The objective has a very small focal length fo and the eyepiece has a small focal length fe, with fo less than fe. The object is placed just outside the focus of the objective. The objective forms a real, magnified, inverted image, which the eyepiece magnifies further. Let L be the distance between the second focus of the objective and the first focus of the eyepiece (the tube length, approximately).

Image at the near point: m = (L/fo)(1 + D/fe).
Final image at infinity: m = (L/fo)(D/fe).

Astronomical (refracting) telescope. The objective has a large focal length fo and a large aperture. The eyepiece has a small focal length fe. In normal adjustment the final image is at infinity:

m = fo/fe (the angle subtended by the image divided by the angle subtended by the object), and the length of the tube is fo + fe.

If the final image is at the near point, m = (fo/fe)(1 + fe/D) and the tube length is fo + ue where ue is the eyepiece object distance.

Reflecting telescope (Cassegrain). A concave primary mirror collects light, and a small convex secondary mirror sends it through a hole in the primary to the eyepiece. Advantages over a refracting telescope:

No chromatic aberration, because mirrors do not disperse light.
A mirror can be supported over its whole back, so it can have a very large aperture and gathers more light.
A paraboloid mirror removes spherical aberration.
The folded path makes the telescope compact.

Instrument	Objective	Eyepiece	Magnifying power
Simple microscope	None (single lens)	Convex lens, small f	1 + D/f (near point); D/f (infinity)
Compound microscope	Convex, very small fo	Convex, small fe	(L/fo)(D/fe) at infinity
Refracting telescope	Convex, large fo and aperture	Convex, small fe	fo/fe in normal adjustment
Reflecting telescope	Concave mirror	Convex lens	Same relation: fo/fe

Formula summary

Topic	Formula	Notes
Mirror focal length	f = R/2	Concave: f, R negative; convex: positive
Mirror equation	1/v + 1/u = 1/f	Cartesian signs, distances in m or cm
Mirror magnification	m = −v/u = h′/h
Refractive index	n = c/v = sin i / sin r
Apparent depth	real depth / n	Viewed normally from a rarer medium
Critical angle	sin ic = 1/n	Denser to rarer medium
Refraction at a spherical surface	n₂/v − n₁/u = (n₂ − n₁)/R
Lens maker	1/f = (n − 1)(1/R₁ − 1/R₂)
Thin lens	1/v − 1/u = 1/f; m = v/u
Power	P = 1/f (f in metre), unit dioptre
Lenses in contact	P = P₁ + P₂; m = m₁m₂
Prism	n = sin((A + δm)/2)/sin(A/2)
Thin prism	δ = (n − 1)A
Simple microscope	1 + D/f; D/f	D = 25 cm
Compound microscope	(L/fo)(D/fe)	Final image at infinity
Telescope	m = fo/fe; length fo + fe	Normal adjustment

Worked examples

Example 1 (concave mirror). An object is placed 15 cm in front of a concave mirror of radius of curvature 20 cm. Find the image.

R = −20 cm, so f = −10 cm. u = −15 cm.

1/v = 1/f − 1/u = −1/10 + 1/15 = −1/30, so v = −30 cm.

m = −v/u = −(−30)/(−15) = −2.

The image is real (v negative), 30 cm in front of the mirror, inverted and twice as large.

Example 2 (convex lens, virtual image). A convex lens has f = +12 cm. An object is placed 8 cm from it. Find the image distance and magnification.

u = −8 cm. 1/v = 1/f + 1/u = 1/12 − 1/8 = −1/24, so v = −24 cm.

m = v/u = (−24)/(−8) = +3.

The image is virtual, erect and three times larger, on the same side as the object.

Example 3 (prism). A prism of angle 60° gives a minimum deviation of 30°. Find its refractive index.

n = sin((60 + 30)/2) / sin(30) = sin 45° / sin 30° = 0.707 / 0.5 = 1.414 (that is, √2).

Example 4 (lens maker). An equiconvex lens of glass with n = 1.5 has radii of curvature 30 cm. Find f and the power.

R₁ = +30 cm, R₂ = −30 cm.

1/f = 0.5 × (1/30 + 1/30) = 0.5 × (1/15) = 1/30, so f = +30 cm.

P = 1/0.30 = +3.33 D.

Example 5 (telescope). A refracting telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. In normal adjustment, find the magnifying power and the length.

m = fo/fe = 100/5 = 20. Length = 100 + 5 = 105 cm.

Example 6 (apparent depth). A tank holds water 1.2 m deep (n = 4/3). The bottom, seen from above, appears at 1.2 / (4/3) = 0.9 m below the surface.

NEET traps

The sign of f: concave mirror f is negative, convex lens f is positive. The Cartesian convention does not use the "real is positive" rule of other conventions.
The mirror equation is 1/v + 1/u, but the lens formula is 1/v − 1/u. Using the wrong one is the most common numerical error.
Mirror magnification is −v/u, while lens magnification is v/u.
Apparent depth is real depth divided by n, not multiplied by n.
TIR is possible only from denser to rarer medium. Light going from air to glass cannot undergo TIR.
Critical angle is smaller for a medium with a larger refractive index (diamond has a smaller ic than glass).
When light enters a new medium, frequency stays constant. Speed and wavelength change.
Power is in dioptre only when f is in metre. A focal length of 20 cm is 0.2 m, so P = 5 D.
In a compound microscope the objective has the smaller focal length. In a telescope the objective has the larger focal length.
Deviation in a prism is minimum when the ray passes symmetrically, not maximum.

One-liners

Focal length of a spherical mirror is f = R/2.
Mirror equation: 1/v + 1/u = 1/f.
Lens formula: 1/v − 1/u = 1/f.
Convex mirror always forms a virtual, erect, diminished image of a real object.
Refractive index n = c/v = sin i / sin r.
Apparent depth = real depth / n.
sin ic = 1/n for a medium to air.
Optical fibres work by total internal reflection.
Diamond sparkles because of its small critical angle.
Lens power P = 1/f in dioptre; powers add for lenses in contact.
Lens maker: 1/f = (n − 1)(1/R₁ − 1/R₂).
Prism: n = sin((A + δm)/2)/sin(A/2).
Thin prism deviation δ = (n − 1)A.
Violet is deviated more than red by a prism.
A telescope in normal adjustment has m = fo/fe and length fo + fe.

Practice questions

Match List I with List II. List I: a. Optical fibre b. Right-angled prism turning a ray by 90° c. Simple microscope d. Cassegrain telescope List II: i. Mirrors as objective ii. Single convex lens of small focal length iii. Repeated total internal reflection in the core iv. Total internal reflection at a face inclined at 45°
1. a-i, b-ii, c-iii, d-iv
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-ii, c-iv, d-i
4. a-iv, b-iii, c-ii, d-i
Answer
B. a-iii, b-iv, c-ii, d-i
a-iii, b-iv, c-ii, d-i.
Match List I with List II. List I: a. Mirror equation b. Thin lens formula c. Lens maker's formula d. Refractive index of a prism List II: i. 1/v + 1/u = 1/f ii. 1/v − 1/u = 1/f iii. 1/f = (n − 1)(1/R₁ − 1/R₂) iv. sin((A + δm)/2)/sin(A/2)
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-iii, b-i, c-ii, d-iv
Answer
A. a-i, b-ii, c-iii, d-iv
Mirror: 1/v + 1/u = 1/f (a-i); thin lens: 1/v − 1/u = 1/f (b-ii); lens maker c-iii; prism d-iv.
A real object is placed in front of a convex mirror. The image is always
1. real, inverted and diminished
2. virtual, erect and enlarged
3. real, inverted and enlarged
4. virtual, erect and diminished
Answer
D. virtual, erect and diminished
A convex mirror forms only a virtual, erect and diminished image of a real object, between P and F behind the mirror.
A glass slab of thickness 6 cm and refractive index 1.5 is placed over a printed letter. Seen from directly above, the letter appears raised by
1. 2 cm
2. 4 cm
3. 6 cm
4. 3 cm
Answer
A. 2 cm
Shift = t(1 − 1/n) = 6(1 − 2/3) = 2 cm. The apparent thickness is 6/1.5 = 4 cm, and the shift is 6 − 4 = 2 cm.
An equiconvex lens of refractive index 1.5 has both radii of curvature equal to 20 cm. Its focal length in air is
1. −20 cm
2. 10 cm
3. 20 cm
4. 40 cm
Answer
C. 20 cm
R₁ = +20, R₂ = −20. 1/f = (1.5 − 1)(1/20 + 1/20) = 0.5 × 0.1 = 0.05 cm⁻¹, so f = +20 cm.
Which of the following statements are correct? 1. The angle of deviation of a prism is the same for every angle of incidence. 2. At minimum deviation, the ray inside an isosceles prism is parallel to its base. 3. Violet light is deviated more than red light by a glass prism.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
Deviation changes with the angle of incidence and is minimum at the symmetric position.
The speed of light in a glass is 2 × 10⁸ m s⁻¹ (c = 3 × 10⁸ m s⁻¹). The refractive index of the glass is
1. 1.33
2. 0.67
3. 2.0
4. 1.5
Answer
D. 1.5
n = c/v = 3 × 10⁸ / 2 × 10⁸ = 1.5.
Two thin lenses of focal lengths +10 cm and −15 cm are placed in contact. The focal length of the combination is
1. +30 cm
2. +6 cm
3. −30 cm
4. +5 cm
Answer
A. +30 cm
1/f = 1/10 − 1/15 = 1/30, so f = +30 cm. Check by power: +10 D − 6.67 D = +3.33 D, so f = 30 cm.
A lens has a power of −2.5 D. It is a
1. concave lens of focal length 40 cm
2. convex lens of focal length 40 cm
3. concave lens of focal length 2.5 m
4. convex lens of focal length 2.5 m
Answer
A. concave lens of focal length 40 cm
Negative power means a diverging lens. f = 1/P = 1/(−2.5) = −0.4 m = −40 cm.
Assertion (A): Large astronomical telescopes use a concave mirror as the objective. Reason (R): A mirror can be supported over its whole back and has no chromatic aberration.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Large mirrors are easier to support and give no colour error.
Match List I with List II. List I: a. Simple microscope, image at near point b. Simple microscope, image at infinity c. Refracting telescope, normal adjustment d. Compound microscope, image at infinity List II: i. fo/fe ii. D/f iii. (L/fo)(D/fe) iv. 1 + D/f
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-iv, c-i, d-iii
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
C. a-iv, b-ii, c-i, d-iii
a-iv, b-ii, c-i, d-iii.
Match List I with List II. List I: a. Lens of f = +50 cm b. Lens of f = −25 cm c. Lens of f = +25 cm d. Lens of f = −50 cm List II: i. +4 D ii. −2 D iii. +2 D iv. −4 D
1. a-iii, b-iv, c-ii, d-i
2. a-i, b-iv, c-iii, d-ii
3. a-i, b-ii, c-iii, d-iv
4. a-iii, b-iv, c-i, d-ii
Answer
D. a-iii, b-iv, c-i, d-ii
P = 1/f in metre: +2 D (a-iii), −4 D (b-iv), +4 D (c-i), −2 D (d-ii).
Light in air is incident on a glass of refractive index √3 at an angle of 60°. The angle of refraction is
1. 90°
2. 30°
3. 60°
4. 45°
Answer
B. 30°
sin r = sin 60° / √3 = (√3/2)/√3 = 1/2, so r = 30°.
An optical fibre has a core of refractive index 1.5 surrounded by a cladding of refractive index 1.2. The sine of the critical angle at the core–cladding boundary is
1. 0.64
2. 0.8
3. 1.25
4. 0.6
Answer
B. 0.8
sin ic = n₂/n₁ = 1.2/1.5 = 0.8 (ic ≈ 53°).
A convex lens of focal length 20 cm forms an image of an object placed 30 cm from the lens. Using the Cartesian convention, the image distance is
1. −60 cm
2. +12 cm
3. +60 cm
4. +30 cm
Answer
C. +60 cm
f = +20 cm, u = −30 cm. 1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60, so v = +60 cm (real image on the other side).
A plano-convex lens has its curved surface of radius 15 cm facing the incident light. The refractive index of the glass is 1.5. Its focal length is
1. +15 cm
2. −30 cm
3. +7.5 cm
4. +30 cm
Answer
D. +30 cm
R₁ = +15, R₂ = ∞. 1/f = (1.5 − 1)(1/15 − 0) = 1/30, so f = +30 cm.
A telescope in normal adjustment has a magnifying power of 20 and its tube length is 84 cm. The focal lengths of the objective and the eyepiece are
1. 20 cm and 1 cm
2. 84 cm and 4.2 cm
3. 80 cm and 4 cm
4. 64 cm and 20 cm
Answer
C. 80 cm and 4 cm
fo/fe = 20 and fo + fe = 84. So 21 fe = 84, fe = 4 cm and fo = 80 cm.
A prism of angle 60° and refractive index √2 is used. The angle of minimum deviation is
1. 45°
2. 60°
3. 15°
4. 30°
Answer
D. 30°
sin((60 + δm)/2) = √2 sin 30° = 0.707, so (60 + δm)/2 = 45° and δm = 30°.
In a compound microscope the objective has focal length 1 cm and the eyepiece has focal length 5 cm. The distance between the second focus of the objective and the first focus of the eyepiece is 20 cm. The magnifying power for the final image at infinity (D = 25 cm) is
1. 500
2. 120
3. 100
4. 4
Answer
C. 100
m = (L/fo)(D/fe) = (20/1)(25/5) = 100. The value 120 is for the final image at the near point: 20 × (1 + 5) = 120.
A convex mirror has a radius of curvature of 40 cm. A real object is placed 20 cm in front of it. The magnification of the image is
1. +0.5
2. −2
3. −0.5
4. +2
Answer
A. +0.5
f = +20 cm, u = −20 cm. 1/v = 1/20 + 1/20 = 1/10, so v = +10 cm. m = −v/u = −10/(−20) = +0.5. Virtual, erect, diminished.
A simple microscope uses a convex lens of focal length 5 cm. With the final image at the near point (D = 25 cm), its magnifying power is
1. 5
2. 6
3. 4
4. 30
Answer
B. 6
m = 1 + D/f = 1 + 25/5 = 6. For the image at infinity it would be 5.
The sparkling of a diamond is mainly due to
1. its large critical angle
2. its small critical angle, which causes repeated total internal reflection
3. its high density
4. absorption of light inside it
Answer
B. its small critical angle, which causes repeated total internal reflection
Diamond has a high refractive index, so its critical angle is small (about 24°). Most light inside undergoes repeated TIR before it comes out.
Assertion (A): A diamond sparkles brightly. Reason (R): Diamond has a small critical angle, so much of the light entering it undergoes repeated total internal reflection.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Small ic leads to repeated TIR inside the stone.
Given below are two statements: Statement I: In a compound microscope the objective has a shorter focal length than the eyepiece. Statement II: In a refracting astronomical telescope the objective has a shorter focal length than the eyepiece.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
In a telescope the objective has the larger focal length so that fo/fe is large.
A pool of water (refractive index 4/3) is 2.4 m deep. Viewed from directly above, the bottom appears at a depth of
1. 3.2 m
2. 2.4 m
3. 1.2 m
4. 1.8 m
Answer
D. 1.8 m
Apparent depth = real depth / n = 2.4 / (4/3) = 1.8 m. Multiplying by n would give the wrong value 3.2 m.
Which of the following statements are correct? 1. A coin under water appears closer to the surface when viewed from above. 2. The apparent depth is the real depth multiplied by the refractive index. 3. A glass slab placed over printed text raises its apparent position.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Apparent depth = real depth divided by n.
A concave mirror of focal length 12 cm forms a real image three times the size of the object. Using the Cartesian convention, the object distance is
1. −8 cm
2. −48 cm
3. −12 cm
4. −16 cm
Answer
D. −16 cm
f = −12, real image so m = −3 and v = 3u. 1/(3u) + 1/u = 1/f gives 4/(3u) = −1/12, so u = −16 cm (v = −48 cm). The value −8 cm gives a virtual image.
Given below are two statements: Statement I: The focal length of a spherical mirror is half its radius of curvature. Statement II: In the Cartesian sign convention, the focal length of a convex mirror is positive.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
f = R/2, and for a convex mirror both f and R are positive.
Given below are two statements: Statement I: The power of a concave lens is positive. Statement II: The power of two thin lenses in contact is the product of their individual powers.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
A concave lens has negative power, and powers of lenses in contact add.
Given below are two statements: Statement I: In normal adjustment the final image of an astronomical refracting telescope is at infinity. Statement II: The length of the telescope in normal adjustment is fo + fe.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard results for normal adjustment.
Match List I with List II. List I: a. Concave mirror b. Convex mirror c. Convex lens d. Concave lens List II: i. f positive, diverging ii. f negative, diverging iii. f negative, converging iv. f positive, converging
1. a-iii, b-i, c-ii, d-iv
2. a-iii, b-i, c-iv, d-ii
3. a-iv, b-i, c-iii, d-ii
4. a-i, b-ii, c-iii, d-iv
Answer
B. a-iii, b-i, c-iv, d-ii
Concave mirror: f negative and converging (a-iii). Convex mirror: f positive and diverging (b-i). Convex lens: f positive, converging (c-iv). Concave lens: f negative, diverging (d-ii).
Given below are two statements: Statement I: The refractive index of a medium is the ratio of the speed of light in vacuum to its speed in the medium. Statement II: When light goes from air into glass, it bends away from the normal.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
On entering the denser medium, light bends toward the normal.
Match List I with List II. List I: a. Speed of light in the medium is 2 × 10⁸ m s⁻¹ (c = 3 × 10⁸ m s⁻¹) b. Critical angle into air is 30° c. Real depth 1.2 m, apparent depth 0.9 m d. Angle of incidence 45°, angle of refraction 30° List II: i. 4/3 ii. 3/2 iii. √2 iv. 2
1. a-ii, b-i, c-iv, d-iii
2. a-i, b-ii, c-iii, d-iv
3. a-iv, b-ii, c-i, d-iii
4. a-ii, b-iv, c-i, d-iii
Answer
D. a-ii, b-iv, c-i, d-iii
n = 1.5 (a-ii); n = 1/sin 30° = 2 (b-iv); n = 1.2/0.9 = 4/3 (c-i); n = sin 45°/sin 30° = √2 (d-iii).
A convex lens of focal length 15 cm is used as a magnifying glass with the object 10 cm from the lens. The magnification of the image is
1. +2
2. +3
3. +1.5
4. −3
Answer
B. +3
f = +15, u = −10. 1/v = 1/15 − 1/10 = −1/30, so v = −30 cm. m = v/u = (−30)/(−10) = +3 (virtual, erect).
Which of the following statements are correct? 1. A convex mirror forms only virtual, erect and diminished images of real objects. 2. A concave mirror forms a virtual image when the object is between its pole and focus. 3. A concave lens forms a real image of a real object.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
A concave lens always forms a virtual image of a real object.
Assertion (A): A glass prism does not split white light into colours. Reason (R): The refractive index of glass depends on the wavelength of light.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Because n depends on wavelength, a prism disperses white light. Assertion A is false.
Assertion (A): A fish in water appears to a viewer directly above to be nearer to the surface than it really is. Reason (R): Apparent depth equals real depth multiplied by the refractive index of water.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Apparent depth = real depth / n, which is smaller than the real depth.
An object 5 cm tall is placed 30 cm from a convex lens of focal length 10 cm. The image formed is
1. real and inverted, 2.5 cm tall
2. real and inverted, 10 cm tall
3. virtual and erect, 2.5 cm tall
4. virtual and erect, 10 cm tall
Answer
A. real and inverted, 2.5 cm tall
u = −30, f = +10. 1/v = 1/10 − 1/30 = 1/15, so v = +15 cm. m = v/u = −0.5, so h′ = −2.5 cm, a real, inverted image.
Which of the following statements are correct? 1. Optical fibres work on total internal reflection. 2. A Cassegrain reflecting telescope has no chromatic aberration. 3. A mirage is formed by total internal reflection of light in air of varying density.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
D. 1, 2 and 3
All three statements are standard facts.
Given below are two statements: Statement I: Total internal reflection can occur when light travels from a rarer to a denser medium. Statement II: The critical angle for a glass–air boundary is larger than that for a diamond–air boundary.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
TIR needs light going from denser to rarer medium. Glass (n = 1.5) has ic ≈ 42° and diamond ic ≈ 24°.
A refracting telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. In normal adjustment its magnifying power is
1. 5
2. 20
3. 0.05
4. 24
Answer
B. 20
m = fo/fe = 100/5 = 20. The value 24 belongs to the near-point adjustment.
Assertion (A): The power of two thin lenses in contact is the sum of their powers. Reason (R): Lens power is measured in dioptre.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
The sum rule follows from 1/f = 1/f₁ + 1/f₂. The unit of power does not explain it.
Assertion (A): A convex mirror is used as a rear-view mirror in vehicles. Reason (R): A convex mirror forms an erect, diminished image and has a wide field of view.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The wide field of view and erect image make it suitable.
When a narrow beam of white light passes through a prism, the colour that is deviated the most is
1. green
2. red
3. violet
4. yellow
Answer
C. violet
The refractive index is larger for shorter wavelengths, so violet is bent the most and red the least.
A concave mirror has a radius of curvature of 40 cm. A real object is placed 60 cm in front of it. Using the Cartesian sign convention, the image distance v is
1. −120 cm
2. +30 cm
3. −30 cm
4. −15 cm
Answer
C. −30 cm
R = −40 cm, f = −20 cm, u = −60 cm. 1/v = 1/f − 1/u = −1/20 + 1/60 = −1/30, so v = −30 cm (real image in front). Check: m = f/(f − u) = −20/40 = −0.5, and −v/u = −(−30)/(−60) = −0.5 agrees.

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1.Which of the following optical instruments creates colourful beautiful patterns due to multiple reflections in plane mirrors?
2.An optical instrument used in a system of prisms, lenses or mirrors to reflect images through a tube is called:
3.The distance between the optical centre of a lens and its principal focus is called ________.
4.Optical fibres work on the principle of ____________.
5.Which of the following can have high optical density but low mass density?
6.Which of these does NOT have an optical lens?
7.Which optical device is made from calcite crystals and is used for observation and production of flat polarized light?
8.Which of the following point(s) always lie(s) on the principal axis of a lens? (i) Point of refraction (ii) Optical Centre (iii) Point of incidence
9.What is the name of the first dedicated Indian astronomy mission that is aimed at studying celestial sources in X-ray, optical and UV spectral bands simultaneously? (RRB NTPC CBT-I (04/02/2021) Shift 1)
10.Travellers in deserts often tend to have an optical illusion of a sheet of water where none actually exists. What is this

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