Two identical glass rods S_1 and S_2 (refract | vedclass

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(B) For the first refraction at the curved surface of $S_1$:Using the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$,where $n_1 = 1.5$,

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) For the first refraction at the curved surface of $S_1$:Using the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$,where $n_1 = 1.5$,$n_2 = 1$,$u = -50 \ cm$,and $R = -10 \ cm$ (as the surface is convex towards the air,the center of curvature is inside the rod,but for the light traveling from glass to air,the surface acts as concave,so $R = -10 \ cm$):$\frac{1}{v} - \frac{1.5}{-50} = \frac{1 - 1.5}{-10}$$\frac{1}{v} + \frac{1.5}{50} = \frac{-0.5}{-10} = 0.05$$\frac{1}{v} = 0.05 - 0.03 = 0.02$$v = 50 \ cm$ (The rays form a virtual image at $50 \ cm$ from the surface of $S_1$ in the air).For the second refraction at the curved surface of $S_2$:The rays must become parallel to the axis inside $S_2$,meaning the image formed by the first surface acts as a virtual object for the second surface at its focal point.Using $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$,where $n_1 = 1$,$n_2 = 1.5$,$v = \infty$,and $R = +10 \ cm$ (convex surface for light entering from air to glass):$\frac{1.5}{\infty} - \frac{1}{-x} = \frac{1.5 - 1}{+10}$$0 + \frac{1}{x} = \frac{0.5}{10} = 0.05$$x = 20 \ cm$ (This is the distance of the virtual object from the surface of $S_2$).Total distance $d = v + x = 50 \ cm + 20 \ cm = 70 \ cm$.

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