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

Answer 2

We are given the following:
Refractive index of the first medium \( n_1 = \frac{4}{3} \),
Refractive index of the second medium \( n_2 = \frac{3}{2} \),
Radius of curvature of the spherical surface \( R = 10 \, \text{cm} \).

The formula for the focal length \( f \) of a spherical surface separating two media is given by the lens-maker's formula for spherical surfaces: \[ \frac{1}{f} = (n_2 - n_1) \left( \frac{1}{R} \right) \] Substituting the given values: \[ \frac{1}{f} = \left( \frac{3}{2} - \frac{4}{3} \right) \times \frac{1}{10} \] First, calculate the difference of the refractive indices: \[ \frac{3}{2} - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \] Now substitute this back: \[ \frac{1}{f} = \frac{1}{6} \times \frac{1}{10} = \frac{1}{60} \] So, the focal length is: \[ f = 60 \, \text{cm} \] Now, using the object distance \( u = -2R = -20 \, \text{cm} \), we use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substituting \( f = 60 \) and \( u = -20 \): \[ \frac{1}{60} = \frac{1}{v} - \frac{1}{-20} \] \[ \frac{1}{60} = \frac{1}{v} + \frac{1}{20} \] Now solve for \( \frac{1}{v} \): \[ \frac{1}{v} = \frac{1}{60} - \frac{1}{20} = \frac{1}{60} - \frac{3}{60} = -\frac{2}{60} = -\frac{1}{30} \] Thus, the image distance is: \[ v = -30 \, \text{cm} \] The negative sign indicates that the image is formed on the same side as the object. This distance is measured in the rarer medium. Thus, the image is formed at a distance of 30 cm in the rarer medium.