Question

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is _____ cm.

Answer 1

Correct Answer: 30

To find the distance of the image from the pole of the refracting surface, we use the formula for refraction at a spherical surface:

(μ₂/v)-(μ₁/u)=(μ₂-μ₁)/R

where:
μ₁ = 1.0 (refractive index of the rarer medium),
μ₂ = 1.5 (refractive index of the denser medium),
u = -15 cm (object distance; negative as per sign convention),
R = +30 cm (radius of curvature; positive as center is on the denser side),
v = image distance from the pole.

Substituting into the formula:

(1.5/v) - (1.0/-15) = (1.5 - 1.0)/30

Let's compute step-by-step:
(1.5/v) + (1.0/15) = 0.5/30

Simplify the right side:
0.5/30 = 1/60

Thus, we have:
(1.5/v) + (1.0/15) = 1/60

Rearrange to find v:
(1.5/v) = 1/60 - 1/15

Find a common denominator for right side:
1/60 - 4/60 = -3/60 = -1/20

Now solve for v:
1.5/v = -1/20

Cross-multiply to find v:
v = -1.5 * 20 = -30 cm

The negative sign indicates the image forms on the opposite side from the incident light. The calculated image distance is |30 cm|, matching the required range 30,30. Thus, the distance of the image from the pole of the surface is 30 cm on the denser side.