Question

Good morning $\huge\boxed{❤️}$

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 sphere ?

No spams please !!!

Answer 1

$\huge\mathfrak{Answer}$

The rays passes through the sphere parallel to the principal axis .[Check attachment for detailed answer]

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¡ Refraction at curves surfaces

It’s formulae :-

n2/v - n1/u = (n2-n1)/R

Refraction of light also have laws.

•It follows snells law

=> sin i/sin r

•Refraction at curved surfaces means on Concave and convex surfaces.

•What is Refraction ?

The bending of light Ray after incidence from one medium to another medium.

•What are the mediums we are talking about ?

-> Rarer medium and Denser Medium

•What happens in rarer medium ?

Refracted ray moves towards the normal

•What happens in Denser medium?

Refracted ray moves away from normal.

<<Refer attachment>>

Answer 2

$\mathfrak{\huge{\blue{\underline{\underline{AnswEr :}}}}}$

Object Distance from the First Surface of the Sphere is $\large{ \frac{R}{(n - 1)} }$

$\mathfrak{\huge{\blue{\underline{\underline{ExplanaTion :}}}}}$

» Given :

❖ Radius of the Sphere = R

❖ Refractive Index of the Sphere = n

❖ u = -x and, v = ∞

❖ n1(Refractive Index of Air) = 1

» To Find :

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 sphere ?

» Solution :

• Using Equation

$\large \boxed{ \frac{n2}{v} - \frac{n1}{u} = \frac{(n2 - n1)}{R} }$

• For the Refraction at first Surface of the Sphere [ Air to Glass ]

» Plugging Values

⇒$\large{ \frac{n}{∞} + \frac{1}{ - x} } = \frac{(n - 1)}{R}$

⇒$\large{ \frac{1}{x} = \frac{(n - 1)}{R} }$

⇒$\huge \boxed{ \red{x = \frac{R}{(n-1)} }}$