JEE Advanced Model Paper 12 with solutions for online practice

From the ray diagram it is clear that ray emerges out of lensparallel to itself. Hence the angle of deviation caused by the lens is zero. From snell’s law we get sin 60° = √3 sin r ⇒ r = 30° . Since the emergent ray is parallel to initial incident ray, the portion of lens used for refraction can be assumed as slab. Hence lateral displacement =

t(sin(i−r))

cosr

=

3sin(60°−30°)

cos30°

= √3 mm . We have , (R−3)2+42 = R2 ⇒ R =

25

6

mm and

1

f

=

(µ−1)

R

⇒ f =

r

(µ−1)

=

25

6(√3−1)

=

25

12

(√3+1) mm