JEE Advance - Physics (2014 - Paper 1 Offline - No. 14)
A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2 = 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then
$$\left| {{f_1}} \right| = 3R$$
$$\left| {{f_1}} \right| = 2.8R$$
$$\left| {{f_2}} \right| = 2R$$
$$\left| {{f_2}} \right| = 1.4R$$
Explanation
As the film has R1 = R2 = R (say), its focal length
$${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$$
$${1 \over f} = 0$$ or $$f = \infty $$. It will not cause refraction.
Refraction will take place at air and glass cylinder. Given u = $$\infty$$
$$ - {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$$
$$ - {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$$ or $${f_1} = 3R$$
For glass-air refraction, we have
$$ - {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$$
$$ - {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$$
or $${f_2} = - 2R$$ or $$\left| {{f_2}} \right| = 2R$$.
$${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$$
$${1 \over f} = 0$$ or $$f = \infty $$. It will not cause refraction.
Refraction will take place at air and glass cylinder. Given u = $$\infty$$
$$ - {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$$
$$ - {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$$ or $${f_1} = 3R$$
For glass-air refraction, we have
$$ - {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$$
$$ - {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$$
or $${f_2} = - 2R$$ or $$\left| {{f_2}} \right| = 2R$$.
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