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Standard XIII

Physics

Lens Formula

The ratio to ...

Question

The ratio to radii of curvature of a biconvex lens $R_{1} : R_{2}$ = 2 : 3. Its refractive index and focal length is $1.5$ and $12 c m$ respectively. What is the value of $R_{1}$ ? Consider $R_{1}$ and $R_{2}$ are positive and negative respectively.

5 c m

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10 c m

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15 c m

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20 c m

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Solution

The correct option is B $10 c m$
We know by lens maker's formula,
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

where $f$ if the focal length, $μ$ is the refractive index and $R_{1}, R_{2}$ are the radii of curvature.

Given,
$\frac{R_{1}}{R_{2}} = \frac{2}{3}$

Take $R_{1}$ = $x$ and $R_{2}$ = $- 3 x / 2$

$\frac{1}{12} = (1.5 - 1) [\frac{1}{x} + \frac{2}{3 x}] \frac{1}{12} = 0.5 [\frac{5}{3 x}] x = \frac{12 \times 2.5}{3} = 10 c m$

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The ratio to radii of curvature of a biconvex lens R1:R2 = 2 : 3. Its refractive index and focal length is 1.5 and 12 cm respectively. What is the value of R1? Consider R1 and R2 are positive and negative respectively.

The correct option is B 10 cmWe know by lens maker's formula, 1f=(μ−1)[1R1−1R2] where f if the focal length, μ is the refractive index and R1, R2 are the radii of curvature. Given, R1R2=23 Take R1 = x and R2 = −3x/2 112=(1.5−1)[1x+23x]112=0.5[53x]x=12×2.53=10 cm

The ratio to radii of curvature of a biconvex lens $R_{1} : R_{2}$ = 2 : 3. Its refractive index and focal length is $1.5$ and $12 c m$ respectively. What is the value of $R_{1}$ ? Consider $R_{1}$ and $R_{2}$ are positive and negative respectively.