Question

A mark placed on the surface of a sphere is viewed through glass from a position directly opposite. If the diameter of the sphere is 10 cm and refractive index of glass is 1.5. The position of the image will be :

• A. -20 cm
• B. 30 cm
• C. 40 cm
• D. -10 cm

Answer 1

The correct option is A -20 cm

The object is placed at 'O'. Its virtual image will be formed by the sphere as shown in the diagram.

Let refraction index of glass $n_1=1.5$

refractive index of air $n_2=1$

Radius of curvature R$=-R=-5cm$

$u=-2R=-10cm$ [from P]

Now, using the formula

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

$\frac{1}{v}-\frac{1.5}{(-10)}=\frac{1-1.5}{-5}$

$\frac{1}{v}+\frac{3}{20}=\frac{1}{10}$

$\frac{1}{v}=\frac{1}{10}-\frac{3}{20}=\frac{-1}{20}$

$v=-20$ [From P]

The position of the image will be $=-20$.