In the figure shown,O is the centre of the gl | vedclass

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(C) When light travels from a denser medium (glass) to a rarer medium (air),it bends away from the normal. For a spherical surface,the refraction form

Vedclass · Accepted Answer

Towards the right of point $O$,shifted towards the observer,i.e.,between $P$ and the observer

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(C) When light travels from a denser medium (glass) to a rarer medium (air),it bends away from the normal. For a spherical surface,the refraction formula is given by $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Here,$n_1 = \mu$ (refractive index of glass),$n_2 = 1$ (refractive index of air),$u = -x$ (where $x$ is the distance of $P$ from the surface),and $R = -R_{sphere}$ (as the surface is concave towards the object). For small angles (normal viewing),the image $I$ is formed closer to the surface than the object $P$. As seen in the provided solution image,the light rays from $P$ refract at the surface and appear to diverge from a point $I$ which is located between $P$ and the surface of the sphere. Therefore,the spot $P$ appears shifted towards the observer,i.e.,between $P$ and the observer.

In the figure shown,$O$ is the centre of the glass sphere. The spot $P$ on the sphere,when viewed almost normally,appears:

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