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  • Shown in the figure is a spherical surface of radius of curvature <img alt="\mathrm{R} \& \mathrm{R} . \mathrm{I} .(=\mathrm{n}) 1.5" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmat

Shown in the figure is a spherical surface of radius of curvature \mathrm{R} \& \mathrm{R} . \mathrm{I} .(=\mathrm{n}) 1.5. The distance of the silvering of the plane surface so as to form an image at the pole due to a very far object.

Option: 1

\mathrm{R}


Option: 2

1.5 \mathrm{R}


Option: 3

2 \mathrm{R}


Option: 4

\mathrm{3 R}


Answers (1)

best_answer

Since light comes from very far object
\mathrm{u}=\infty
\Rightarrow \frac{\mathrm{n}}{\mathrm{v}}-\frac{1}{\infty}=\frac{\mathrm{n}-1}{\mathrm{R}}
\Rightarrow \mathrm{v}=\frac{\mathrm{nR}}{\mathrm{n}-1}=\frac{1.5 \mathrm{R}}{1.5-1}=3 \mathrm{R}.


Due to presence of the plane mirror, the image found at I behaves as a virtual object for the plane mirror & a real image I’ is formed in front of the plane mirror, at the pole P.

\Rightarrow x+x=v
\mathrm{ \Rightarrow x=\frac{v}{2}=\frac{3 R}{2} }
\mathrm{ \Rightarrow \quad x=1.5 R }.

Posted by

Kuldeep Maurya

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