Question

A spherical surface of radius 30 cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?

Answer 1

Answer:

a is better

Explanation:

the answer is a

Answer 2

At a distance of 266 cm a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface.

Explanation:

For par-axial rays after refraction are parallel. The Image is forever created

Refraction :

When a beam of light goes over from one transparent medium to another, Crossing the screen it bends, separating the two outlets. This is called the refraction

Given data in the question  :

v = ∞    

$\mu_{1}$ = 1.33

u = ?

$\mu_{2}$ = 1.48

R = 30cm ⇒ Curvature Radius

We know that,

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

Put the values in above equation

$\frac{1.48}{\infty}-\frac{1.33}{u}=\frac{1.48-1.33}{30}$

$\frac{1.48}{\infty}-\frac{1.33}{u}=\frac{0.15}{30}$

       $-\frac{1.33}{u}=\frac{0.15}{30}$

$-1.33 \times 30=0.15 u$

               $u=-\frac{1.33 \times 30}{0.15}$

               $u=-\frac{39.9}{0.15}$

                $u=-266$

Thus, the subject should be put on side A at a distance of 266 cm from the (convex) surface.