Question

36. A convex lens of focal length 0.24m and of refractive index 1.5 is completely immersed in
water of refractive index 1.33. Find the change in the focal length of the lens.​

Answer 1

The change in focal length of the lens is 0.6975 m or 69.75 cm

Explanation:

Given Data

A Convex lens have,

Focal length = 0.24 m = 24 cm

Refractive Index = 1.5 ( glass )

Refractive Index = 1.33 ( water )

Refractive index = 1.0003 = 1 (air)

Find the change in focal length of the lens.

The lens makers formula is

$\frac{1}{f}=\left[\frac{n_{2}}{n_{1}}-1\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

In air,

$\frac{1}{f_{\text {air }}}=\left[\frac{n_{2}}{n_{1}}-1\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{24}=\left[\frac{1.5}{1}-1\right]\left[\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right]$

$\frac{1}{24}=0.5\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]--------(1)$

In water,

$\frac{1}{f_{\text {water }}}=\left[\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}-1\right]\left[\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right]$

$\frac{1}{f_{\text {water }}}=\left[\frac{1.5}{1.33}-1\right]\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{f_{\text {water }}}=0.128\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]--------(2)$

Dividing (1) by (2) we get,

$\frac{f_{\text {water }}}{24}=\frac{0.5}{0.128}$

$f_{\text {water }}=\frac{0.5}{0.128} \times 24$

$f_{\text {water }}= 93.75 \mathrm{cm}$

In water, the focal length is 93.75 cm.

Change in focal length of the lens = Focal length in water - focal length in air

Change in focal length of the lens = 93.75 - 24

Change in focal length of the lens = 69.75 cm

Change in focal length of the lens = 0.6975 m

Therefore, the change in focal length of the lens is 0.6975 m or 69.75 cm when the focal length of lens in water and air are 93.75 cm (0.9375 m) and 0.24 m and the refractive index of the lens in water and air are 1.5 and 1.33.

To Learn more ...

1) How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia? Explain it mathematically.

https://brainly.in/question/1133883

2) The refractive index of a lens material is 1.5 and focal length f due to some chemical changes in the material its refractive index has increased by 2% the percentage change in its focal length is

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