10A. Solved Numerical Examples
Free study material · concepts, shortcuts & solved questions
A small number of numerical questions on mirrors and lenses do appear in SSC/RRB General Science and General Intelligence sections, usually testing direct substitution into the mirror or lens formula. The following worked examples illustrate the method.
Example 1: Concave mirror
An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the position and nature of the image.
Solution: By sign convention, u = −15 cm and f = −10 cm (concave mirror focal length is negative). Using 1/v + 1/u = 1/f: 1/v = 1/f − 1/u = (−1/10) − (−1/15) = −1/10 + 1/15 = (−3 + 2)/30 = −1/30, so v = −30 cm. Since v is negative, the image is formed 30 cm in front of the mirror (real side), so it is a real, inverted image. Magnification m = −v/u = −(−30)/(−15) = −2, meaning the image is inverted and twice the size of the object (magnified).
Example 2: Convex lens
An object is placed 20 cm from a convex lens of focal length 15 cm. Find the image distance and magnification.
Solution: By sign convention, u = −20 cm and f = +15 cm (convex lens focal length is positive). Using 1/v − 1/u = 1/f: 1/v = 1/f + 1/u = 1/15 + (−1/20) = (4 − 3)/60 = 1/60, so v = 60 cm. The positive value of v means the image is formed on the side opposite to the object — a real, inverted image. Magnification m = v/u = 60/(−20) = −3, so the image is inverted and three times the size of the object.
Example 3: Power of a lens
An optician prescribes a lens of power −2.5 D to a patient. Identify the type of lens, the defect being corrected, and its focal length.
Solution: Since the power is negative, the lens is a concave (diverging) lens, prescribed to correct myopia (short-sightedness). Focal length f = 1/P = 1/(−2.5) = −0.4 m = −40 cm (the negative sign confirms it is a concave lens, consistent with the sign convention).
Example 4: Refractive index and critical angle
Calculate the critical angle for a glass-air boundary if the refractive index of the glass is 1.5.
Solution: Using sin C = 1/n = 1/1.5 = 0.667, so C = sin⁻¹(0.667) ≈ 41.8°, which matches the commonly quoted critical angle of about 42° for ordinary glass.