. A transparent sphere of radius ‘y’ 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?
Answers
Answer:
The object should be placed at the distance of \frac{R}{\mu - 1}
μ−1
R
from the surface of the sphere.
Explanation:
Radius of sphere = R
Refractive Index of the sphere = \muμ
Using Equation
\frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_2-\mu_1}{R}
v
μ
1
−
u
μ
2
=
R
μ
2
−μ
1
For the refraction at the first surface of the sphere.(Air to glass)
\frac{\mu}{\infty}+\frac{1}{x} = \frac{\mu - 1}{R}
∞
μ
+
x
1
=
R
μ−1
[v = ∞, \mu_2 = \muμ
2
=μ , u = -x, \mu_1μ
1
= 1]
\frac{\mu}{\infty} + \frac{1}{x} = \frac{\mu - 1}{R}
∞
μ
+
x
1
=
R
μ−1
⇒\frac{1}{x} = \frac{\mu - 1}{R}
x
1
=
R
μ−1
⇒x = \frac{R}{\mu - 1}x=
μ−1
R
Hence,the object should be placed at the distance of \frac{R}{\mu - 1}
μ−1
R
from the surface of the sphere to obtain the real image at the distance from the second surface of the sphere.