A transparent thin film of uniform thickness | vedclass

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(B) The thin film has uniform thickness,so its surfaces are parallel. The power of such a film is $\frac{1}{f_{	ext{film}}} = (n_1 - 1) \left( \frac{

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$(A, C)$

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(B) The thin film has uniform thickness,so its surfaces are parallel. The power of such a film is $\frac{1}{f_{	ext{film}}} = (n_1 - 1) \left( \frac{1}{R} - \frac{1}{R} ight) = 0$,which means $f_{	ext{film}} = \infty$. Thus,the film does not change the direction of the parallel rays.From Air to Glass:Using the formula for refraction at a single spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.Here,$n_1 = 1$ (air),$n_2 = 1.5$ (glass),$u = \infty$,and the radius is $R$. $\frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5 - 1}{R} \Rightarrow \frac{1.5}{v} = \frac{0.5}{R} \Rightarrow v = 3R$.So,$|f_1| = 3R$.From Glass to Air:Using the formula for refraction at a single spherical surface: $\frac{n_1}{v} - \frac{n_2}{u} = \frac{n_1 - n_2}{-R}$.Here,$n_1 = 1$ (air),$n_2 = 1.5$ (glass),$u = \infty$,and the radius is $-R$ (as the surface is concave from the glass side).$\frac{1}{v} - \frac{1.5}{\infty} = \frac{1 - 1.5}{-R} \Rightarrow \frac{1}{v} = \frac{-0.5}{-R} \Rightarrow v = 2R$.So,$|f_2| = 2R$.Therefore,statements $(A)$ and $(C)$ are correct.

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