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  • The focal length of a convex lens of R.I. 1.5 is when it is placed in air. When it is immersed in a l

The focal length of a convex lens of R.I. 1.5 is \mathrm{f} when it is placed in air. When it is immersed in a liquid its focal length becomes \mathrm{xf}. The refractive index of the liquid

Option: 1

>3 / 2


Option: 2

<(3 / 2)$ and $>1


Option: 3

<3 / 2


Option: 4

All of these


Answers (1)

best_answer

1 / \mathrm{f}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right)
\Rightarrow \quad 1 / \mathrm{f}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right) \text { when the lens is placed in air and }
\mathrm{1/xf}=\left(\frac{1.5}{\mathrm{y}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right) \text { when the lens is places in the liquid}

Where \mathrm{y=R . I}\mathrm{y=R . I.}  of the liquid

\mathrm{\Rightarrow \frac{f}{x f}=\frac{\frac{1.5}{y}-1}{1.5-1}}

\mathrm{\Rightarrow 1+\frac{1}{2 x}=\frac{1.5}{y}}
\mathrm{\Rightarrow y=\frac{3}{2}\left(\frac{2 x}{2 x+1}\right)}\mathrm{y= \frac{3}{2}\left ( \frac{2x}{2x+1} \right )}
\mathrm{\Rightarrow y=\frac{3}{2}\left(\frac{2 x}{2 x+1}\right)}\mathrm{\Rightarrow y=\frac{3}{2}\left(\frac{2 x}{2 x+1}\right)}
\mathrm{\Rightarrow \quad y=\frac{3 x}{2 x+1}}

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shivangi.bhatnagar

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