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 requires calculating the image distance formed by a spherical refracting surface using the spherical refraction formula: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} \). The following parameters are given:

• \( n_1 = 1 \) (refractive index of air)
• \( n_2 = \frac{4}{3} \) (refractive index of the material)
• \( u = -20 \) (object distance, negative by convention)
• \( R = 10 \) (radius of curvature, positive for a convex surface from air)

Substituting these values into the formula yields:

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

The calculation proceeds as follows:

• \( \frac{4}{3v} = \frac{1}{20} + \frac{1}{30} \)
• \( \frac{4}{3v} = \frac{1}{20} + \frac{1}{30} \)
• Combining terms on the right side: \( \frac{1}{20} = \frac{3}{60} \) and \( \frac{1}{30} = \frac{2}{60} \). Therefore, \( \frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12} \)
• This simplifies the equation to: \( \frac{4}{3v} = \frac{1}{12} \)
• Cross-multiplication results in: \( 4 \times 12 = 3v \), which gives \( v = \frac{48}{3} = 16 \, \text{cm} \)

The positive image distance (v = 16 cm) signifies that the image is formed in air, on the same side as the object, at a distance of 16 cm from point P.