Solve

Guides

Question

A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The center of curvature is in glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to

Open in App

Solution

Verified by Toppr

Let us assume

OP = PQ = a (say)

According to sign conventions,

Object distance (u) = -x

Image distance (v) = +x

We know the formula

μ2v - μ1u = μ2 - μ1R

Substituting the values in the above formula,

1.5x - 1 - x = 1.5 - 1R

1.5x + 1x = 0.5 R

2.5x = 0.5 R

x = 5R

The distance PO is equal to 5R.

Was this answer helpful?

0