Question

In the figure given below, APB is a curved surface of radius of curvature 10 cm separating air and a transparent material \((μ = \frac{4}{3} ).\) A point object O is placed in air on the principal axis of the surface 20 cm from P. The distance of the image of O from P will be _____.
Fill in the blank with the correct answer from the options given below

• A. 16 cm left of P in air
• B. 16 cm right of P in water
• C. 20 cm right of P in water
• D. 20 cm left of P in air

Answer 1

The Correct Option is A

The problem involves determining the distance of the image formed by a spherical refracting surface using the formula for refraction at a spherical surface. The formula is given by: 
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}\), where:

• \(n_1 = 1\) (refractive index of air)
• \(n_2 = \frac{4}{3}\) (refractive index of the material)
• \(u = -20 \)\) (object distance, negative according to sign convention)
• \(R = 10 \)\) (radius of curvature, positive as it is measured from air to the convex surface)

Substitute these values into the formula:

\(\frac{\frac{4}{3}}{v} - \frac{1}{-20} = \frac{\frac{4}{3}-1}{10}\)

Solving step by step:

• \(\frac{4}{3v} = \frac{1}{20} + \frac{1}{30}\)
• \(\frac{4}{3v} = \frac{1}{20} + \frac{1}{30}\)
• Finding common denominator for right side:
  \(\frac{1}{20} = \frac{3}{60}\) and \(\frac{1}{30} = \frac{2}{60}\), so \(\frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}\)
• Thus, \(\frac{4}{3v} = \frac{1}{12}\)
• Cross-multiplying gives: \(4 \times 12 = 3v \Rightarrow v = \frac{48}{3} = 16 \, \text{cm}\)

The positive value of v indicates that the image is formed in air on the same side as the object, hence 16 cm left of P in air.