Question

The diameter of a glass sphere is 10 cm a beam of light strikes the surface of sphere which converges at 20 cm from the pole of sphere surface find position of the final image ​

Answer 1

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Given the diameter of class 10 cm. A beam of light strikes the sphere which converges at a point 20 cm behind the pole of spherical surface. Find position of image of mu equals to 1.5.

We are given v = + 30 (sphere that converges behind the pole of spherical surface) .We know that μ2 / v – μ1 / u = μ2 – μ1 / R

$\frac{1.5}{v} - \frac{1}{30} = \frac{2}{5} \times \frac{1}{2} (r = \frac{15}{2} ) \\ \frac{1.5}{v} - \frac{1}{30} = \frac{1}{5} \\ \frac{1.5}{v} + \frac{1}{30} \\ \frac{1.5}{v} = 1 + \frac{2}{30} \\ \frac{1.5}{v} = \frac{1}{10} \\ so \: v = 10 \times 1.5 \\ so \: v = 15$

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Answer 2

Answer:

The position of the final image from the pole of the sphere surface is 30cm.

Explanation:

We will use the following formula to solve this question. Which is given as,

$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$      (1)

Where,

μ₁=the medium in which the light is present

μ₂=the medium in which the light has to travel

v=image distance

u=object distance

R=radius of the sphere

From the question we have,

μ₁=1       (considering medium 1 as air)

μ₂=1.5   (as the sphere is made up of glass)

The diameter of the sphere=10cm

So, the radius is=5cm

The object distance=20cm

By substituting the values in equation (1) we get;

$\frac{1.5}{v}-\frac{1}{-20}=\frac{1.5-1}{5}$

$\frac{1.5}{v}+\frac{1}{20}=\frac{1.5-1}{5}$

$\frac{1.5}{v}+\frac{1}{20}=\frac{0.5}{5}$

$\frac{1.5}{v}+0.05=0.1$

$\frac{1.5}{v}=0.1-0.05$

$\frac{1.5}{v}=0.1-0.05$

$\frac{1.5}{v}=0.05$

$v=\frac{1.5}{0.05}$

$v=30cm$

As a result, the final image's distance from the sphere's pole is 30cm.

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