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(D) The light rays from point $A$ on the wall travel through the glass and refract at the spherical surface into the air.For refraction at a spherical

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(D) The light rays from point $A$ on the wall travel through the glass and refract at the spherical surface into the air.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 glass $(\mu_1 = 1.5)$ to air $(\mu_2 = 1.0)$.The object distance $u$ is the distance of $A$ from the pole $X$. Since $XA = 3 	ext{ cm}$ and light travels from $A$ to $X$, $u = -3 	ext{ cm}$.The radius of curvature $R$ for the convex surface (as seen from the glass side) is $-5 	ext{ cm}$ because the center of curvature lies to the left of the pole $X$.Substituting these values into the formula:$\frac{1.0}{v} - \frac{1.5}{-3} = \frac{1.0 - 1.5}{-5}$$\frac{1}{v} + 0.5 = \frac{-0.5}{-5} = 0.1$$\frac{1}{v} = 0.1 - 0.5 = -0.4$$v = -\frac{1}{0.4} = -2.5 	ext{ cm}$.The negative sign indicates that the image $I$ is formed $2.5 	ext{ cm}$ to the left of the pole $X$.The distance of the observer $O$ from the pole $X$ is $OX = OA - XA = 8 - 3 = 5 	ext{ cm}$.The apparent distance of $A$ from the observer $O$ is $OI = OX + XI = 5 	ext{ cm} + 2.5 	ext{ cm} = 7.5 	ext{ cm}$.

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