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  •  A small filament is at the centre of a hollow glass sphere of inner and outer radii

 A small filament is at the centre of a hollow glass sphere of inner and outer radii 8 \mathrm{~cm} and 9 \mathrm{~cm} respectively. The refractive index of glass is 1.50 . Calculate the position of the image of the filament when viewed from outside the sphere.

Option: 1

9 \mathrm{~cm}


Option: 2

-9 \mathrm{~cm}


Option: 3

-19 \mathrm{~cm}


Option: 4

+19 \mathrm{~cm}


Answers (1)

For refraction at the first surface,

\mathrm{u}=-8 \mathrm{~cm}, \mathrm{R}_{1}=-8 \mathrm{~cm}, \mu_{1}=1, \mu_{2}=1.5

\mathrm{\frac{m_{2}}{v}-\frac{m_{1}}{u}=\frac{m_{2}-m_{1}}{R_{1}}}
\mathrm{\frac{1.5}{v^{\prime}}+\frac{1}{8}=\frac{0.5}{-8}}
\mathrm{\mathrm{v}^{\prime}=-8 \mathrm{~cm}}


It means due to the first surface the image is formed at the centre. For the second surface

\mathrm{\mathrm{u}=-9 \mathrm{~cm}, \mu_{1}=1.5, \mathrm{R}_{2}=-9 \mathrm{~cm}}
\frac{\mathrm{m}_{2}}{\mathrm{v}}-\frac{\mathrm{m}_{1}}{\mathrm{u}}=\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{\mathrm{R}_{2}}
\frac{1}{\mathrm{~v}}+\frac{1.9}{5}=\frac{1-1.5}{-9}
v=-9 \mathrm{~cm}

Thus, the final image is formed at the centre of the sphere.
 

Posted by

Kshitij

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