Question

For the spherical interface shown in the figure, the two different media with refractive indices$n_1=1.4$and $n_2=1.25$are present as shown. The image will be formed at

• A. $-\frac{125}{3}cm$
• B. $-\frac{50}{6}cm$
• C. $-\frac{25}{2}cm$
• D. $-20cm$

Answer 1

The correct option is A

$-\frac{125}{3}cm$

Step 1. Given data:

Taking into account the sign convention the object position and radius of the interface will be negative

Position of the object, $u=-40cm$

Refractive index of medium 1,$n_1=1.4$

Refractive index of medium 2, $n_2=1.25$

The radius of the interface, $R=-30cm$

Step 2. Finding the image distance:

For a Spherical interface,

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

Putting the values, we get

$\frac{1.25}{v}-\frac{1.4}{-40}=\frac{1.25-1.4}{-30}$

$\Rightarrow \frac{1.25}{v}=\frac{-1.25+1.4}{30}-\frac{1.4}{40}$

$\Rightarrow \frac{1.25}{v}=\frac{0.15}{30}-\frac{1.4}{40}$

$\Rightarrow \frac{1.25}{v}=0.005-0.035=-0.03$

$\Rightarrow v=-\frac{1.25}{0.03}=-\frac{125}{3}cm$, the negative value signifies that the image is formed in the same side as object and is virtual.

Hence, option A is correct