Question

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is _____ cm.

Answer 1

Correct Answer: 30

The refraction at a spherical surface is governed by the equation: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}, \] where: 
\( \mu_1 = 1.0 \) (refractive index of the rarer medium), 
\( \mu_2 = 1.5 \) (refractive index of the denser medium), 
 \( u = -15 \, \text{cm} \) (object distance, negative as it is measured opposite to the direction of incident light), 
 \( R = +30 \, \text{cm} \) (radius of curvature, positive as the center of curvature lies in the denser medium), 
 \( v \) is the image distance to be determined. 
 Substitute the values into the equation: \[ \frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}. \] 
Simplify: \[ \frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}. \] \[ \frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}. \] 
Simplify further: \[ \frac{1.5}{v} = \frac{1 - 4}{60} = \frac{-3}{60}. \] \[ \frac{1.5}{v} = -\frac{1}{20}. \] Solve for \( v \): \[ v = -\frac{1.5 \cdot 20}{1} = -30 \, \text{cm}. \] The negative sign indicates that the image is formed on the opposite side of the refracting surface, i.e., in the denser medium. 
Final Answer: The distance of the image from the pole of the surface is: \[ \boxed{30 \, \text{cm}}. \]