Question

A transparent sphere of radius R and refractive index n is kept in air. At what distance from the surface of the sphere should a point object be placed on the principal axis so as to form a real image at the same distance from the second surface of the sphere ?solve the problem spammers get lost​

Answer 1

The object should be placed at the distance of $\frac{R}{\mu - 1}$ from the surface of the sphere.

Explanation:

Radius of sphere = R

Refractive Index of the sphere = $\mu$

Using Equation

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

For the refraction at the first surface of the sphere.(Air to glass)

$\frac{\mu}{\infty}+\frac{1}{x} = \frac{\mu - 1}{R}$      [v = ∞, $\mu_2 = \mu$, u = -x, $\mu_1$ = 1]

$\frac{\mu}{\infty} + \frac{1}{x} = \frac{\mu - 1}{R}$

⇒$\frac{1}{x} = \frac{\mu - 1}{R}$

⇒$x = \frac{R}{\mu - 1}$

Hence,the object should be placed at the distance of $\frac{R}{\mu - 1}$ from the surface of the sphere to obtain the real image at the distance from the second surface of the sphere.

Answer 2

Given that,

Radius of the sphere = R

Refractive index of the medium = n

Let the distance is x

We need to calculate the object distance

Using formula of refractive index

$\dfrac{n_{2}}{v}+\dfrac{n_{1}}{u}=\dfrac{n_{2}-n_{1}}{r}$

Where, n = refractive index

u = object distance

v = image distance

r = radius of sphere

Put the value into the formula

$\dfrac{n}{\infty}+\dfrac{1}{x}=\dfrac{n-1}{R}$

$x=\dfrac{R}{n-1}$

Hence, The object will be placed at the distance of $\dfrac{R}{n-1}$ from the surface of the sphere to obtain the real image at the distance from the second surface of the sphere.