Question

A biconvex lens of refractive index 1.5 has a focal length 10cm. Calculate the
radius of curvature. Find the position and nature of the image of an object held at a
distance of 10 cm from the lens.​

Answer 1

(a). The radius of curvature is 10 cm.

(b). The image distance is ∞

Explanation:

Given that,

Refractive index = 1.5

Focal length = 10 cm

Object distance = 10 cm

For biconvex lens,

$R=R_{1}=-R_{2}$

We need to calculate the radius of curvature

Using formula of radius of curvature

$\dfrac{1}{f}=(\mu-1)(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{10}=(1.5-1)(\dfrac{1}{R}+\dfrac{1}{R})$

$\dfrac{1}{10}=(1.5-1)\times\dfrac{2}{R}$

$R=(1.5-1)\times20\ cm$

The radius of curvature is 10 cm.

We need to calculate the image distance

Using formula of  lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{10}=\dfrac{1}{v}+\dfrac{1}{10}$

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{10}$

$v =\infty$

Hence, (a). The radius of curvature is 10 cm.

(b). The image distance is ∞

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Topic : optics

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