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The graph between $\mathrm{\frac{1}{u}}$ and $\mathrm{\frac{1}{v}}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $\mathrm{1.5}$ and its both the surfaces have same radius of curvature $\mathrm{R. }$ The value of $\mathrm{R}$ will be _________ $\mathrm{\mathrm{cm}.}$

$\mathrm{\text { (where } u=\text { object distance, } v=\text { image distance) }}$

Option: 1
10

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

By lens maker formula,
$\mathrm{ \frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) }$

$\mathrm{ R=R_1=R_2, \mu=1.5 }$

$\mathrm{ \therefore \frac{1}{f} =(\mu-1) \times \frac{2}{R_1}=\frac{1}{R} }$

Or

$\mathrm{\frac{1}{f}=(\mu-1) \times \frac{2}{R_2}=\frac{1}{R}}$

From thr graph,

For, $\mathrm{\frac{1}{\mu }=-0.10}$

$\mathrm{\frac{1}{v}=-0}$

$\mathrm{\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=\frac{1}{R} }$

$\mathrm{0-(0.1)=\frac{1}{R}}$

$\mathrm{0 \cdot1=\frac{1}{R}}$

$\mathrm{\Rightarrow R=10\: cm}$

The value of R will be $\mathrm{10 \mathrm{~cm}}$

Posted by

vishal kumar

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