JEE Advance - Physics (2014 - Paper 2 Offline - No. 11)
A point source S 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 below 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
1.21
1.30
1.36
1.42
Explanation
In $$\Delta$$SAB
$$\tan \theta = {{AB} \over {AS}} = {{11.54/2} \over {10}}$$
tan$$\theta$$ = 0.577
$$\theta$$ = 30$$^\circ$$
Using $${{\sin {\theta _l}} \over {\sin 90}} = {{{n_l}} \over {{n_B}}}$$, we get
$$\sin {\theta _l} = {{{n_l}} \over {{n_B}}}$$ or $${n_l} = {n_B} \times \sin 30 = 2.72 \times {1 \over 2} = 1.36$$
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