Test: Refraction by Spherical Surfaces - JEE MCQ

Solution : 

The correct option is Option A.

Laws of refraction;-

The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.

For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.

When a ray of light passes through a glass slab, ∠i,∠r and the normal all lie in the same plane.

When a ray of light passes from one medium to another, here from air to glass or glass to air, the ratio sini / sinr = constant.

For plane mirror R=infinity
so. (n₂-n₁)/R=0
therefore, n₂/V=n₁/U

The relation between radius of curvature, image distance and object distance is given by
n₂/v-n₁/u=n₂-n₁
un₂-vn₂/vu=n₂-n₁/R
R=vu(n2-n2)/un₂-vn₂
=uv(2)/u(2)-v(1)
=[uv/2u-v)

The image will be virtual, upright and diminished.

A thick Plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face, and we have to determine the value of ' x'
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm