In a thin spherical fish bowl of radius 10 , | vedclass

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(A) To find the position of the image,we use the formula for refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu

Vedclass · Accepted Answer

$5.2$

Vedclass · Answer

(A) To find the position of the image,we use the formula for refraction at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travels from water to air,so $\mu_1 = \frac{4}{3}$ and $\mu_2 = 1$.The radius of the spherical surface is $R = -10 \, cm$ (using sign convention,as the center is to the left of the surface).The object (fish) is at a distance of $4 \, cm$ from the center $C$ towards the right. Since the radius is $10 \, cm$,the distance of the fish from the surface $E$ is $u = -(10 - 4) = -6 \, cm$.Substituting these values into the formula:$\frac{1}{v} - \frac{4/3}{-6} = \frac{1 - 4/3}{-10}$$\frac{1}{v} + \frac{4}{18} = \frac{-1/3}{-10}$$\frac{1}{v} + \frac{2}{9} = \frac{1}{30}$$\frac{1}{v} = \frac{1}{30} - \frac{2}{9} = \frac{3 - 20}{90} = \frac{-17}{90}$$v = -\frac{90}{17} \approx -5.29 \, cm$.The negative sign indicates that the image is formed on the same side as the object,at a distance of $5.29 \, cm$ from the surface $E$. Rounding to the nearest given option,the answer is $5.2 \, cm$.

In a thin spherical fish bowl of radius $10 \, cm$ filled with water of refractive index $\mu = \frac{4}{3}$,there is a small fish at a distance of $4 \, cm$ from the centre $C$ as shown in the figure. Where will the image of the fish appear if seen from $E$ (in $, cm$)?

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