Skip to content
dronacharya.ai
← JEE Advanced 2014 · Paper 2
PhysicsSingle correct+3 / -1Question 5 of 60

JEE Advanced 2014 · Paper 2, Physics question 5

A point source SS is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

A rectangular transparent block resting at the bottom of a container of liquid, with a point source S at the middle of its bottom face; rays from S go up to the top face of the block, which is covered by the liquid.
Choose one option
Answer and worked solution

Official answer: C

Edge of the spot. Light escapes only within the critical angle. The spot radius is 11.54/2=5.7711.54/2=5.77 mm at height 10 mm:
tan⁡θc=5.7710=0.577  ⟹  θc=30°.\tan\theta_c=\frac{5.77}{10}=0.577\implies\theta_c=30°.

Critical angle. sin⁡θc=nℓ2.72\sin\theta_c=\dfrac{n_\ell}{2.72}, so
nℓ=2.72×12=1.36.n_\ell=2.72\times\frac12=1.36.

(C)

Topics:Total internal reflection
← Previous questionNext question →