A parallel beam of light travelling in water | vedclass

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Question

(A) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travels from wa

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$6 \, cm$ from the first surface

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(A) For refraction at a spherical surface,the formula is: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travels from water $(\mu_1 = 4/3)$ to air $(\mu_2 = 1)$.The incident beam is parallel,so the object distance $u = \infty$.The radius of curvature $R$ for the first surface (convex towards the incident light) is $-2 \, cm$ (using sign convention,as the center of curvature is to the left of the surface).Substituting the values: $\frac{1}{v} - \frac{4/3}{\infty} = \frac{1 - 4/3}{-2}$.$\frac{1}{v} - 0 = \frac{-1/3}{-2} = \frac{1}{6}$.Therefore,$v = 6 \, cm$.The image is formed at $6 \, cm$ from the first surface in the direction of light propagation.

$A$ parallel beam of light travelling in water (refractive index $\mu_1 = 4/3$) is refracted by a spherical air bubble of radius $R = 2 \, cm$ situated in water. Assuming the light rays to be paraxial,the position of the image due to refraction at the first surface is:

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Region $I$ and $II$ are separated by a spherical surface of radius $25\, \text{cm}$. An object is kept in region $I$ at a distance of $40\, \text{cm}$ from the surface. The distance of the image from the surface is $.....\, \text{cm}$.

Explain the refraction by two transparent spherically curved surfaces.

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