Question

A luminous point object O is placed at a distance 2R from the spherical boundary separating two transparent media of refractive indices n₁ and n₂ as shown, where R is the radius of curvature of the spherical surface. If \(n_1=\frac{4}{3}, n_2=\frac{3}{2}\) and R =10 cm, the image is obtained at a distance from P equal to

• A. 30 cm in the rarer medium
• B. 30 cm in the denser medium
• C. 18 cm in the rarer medium
• D. 18 cm in the denser medium

Answer 1

The Correct Option is A

Given:
\(n_1 = \frac{4}{3}\) (refractive index of the rarer medium)
\(n_2 = \frac{3}{2}\) (refractive index of the denser medium)
\(R = 10 \, \text{cm}\) (radius of curvature of the spherical surface) The object is at a distance \(2R = 20 \, \text{cm}\) from the spherical boundary. The formula for the image formed by a spherical surface separating two media is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] Where: - \(u = -2R = -20 \, \text{cm}\) (object distance) - \(v\) is the image distance, which we need to find. Substitute the known values: \[ \frac{3/2}{v} - \frac{4/3}{-20} = \frac{\frac{3}{2} - \frac{4}{3}}{10} \] Simplifying the equation: \[ \frac{3/2}{v} + \frac{4}{60} = \frac{9 - 8}{30} \] \[ \frac{3/2}{v} + \frac{1}{15} = \frac{1}{30} \] Now, solve for \(v\): \[ \frac{3/2}{v} = \frac{1}{30} - \frac{1}{15} = -\frac{1}{30} \] \[ v = \frac{-3/2}{-1/30} = 30 \, \text{cm} \] So, the image is formed at a distance of 30 cm in the rarer medium.
Thus, the correct answer is: Correct Answer: (A) 30 cm in the rarer medium