Question

Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.

Hint:

Apply the formula for refraction at a spherical surface, \( \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \). Be careful with the sign conventions for \( u \), \( v \), and \( R \). For a convex surface, \( R \) is positive when light goes from a rarer to a denser medium. Real objects have negative \( u \), and real images formed on the opposite side of the refracting surface have positive \( v \). Ensure consistent units throughout the calculation and convert the final answer to the required unit.

Answer 1

Correct Answer: 4

Utilizing the spherical surface refraction formula: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]. Parameters are: \( \mu_1 \) (refractive index of medium 1, air) = 1; \( \mu_2 \) (refractive index of medium 2, glass) = 1.5; \( u \) (object distance from surface, to be determined); \( v \) (image distance from surface) = 200 cm (positive for image in medium 2); \( R \) (radius of curvature of surface) = +50 cm (positive for convex surface relative to incident light). Substitution yields: \[ \frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50} \] \[ \frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} \] \[ \frac{1.5}{200} - \frac{1}{u} = \frac{1}{100} \] \[ \frac{3}{400} - \frac{1}{u} = \frac{1}{100} \] \[ \frac{1}{u} = \frac{3}{400} - \frac{1}{100} = \frac{3}{400} - \frac{4}{400} = -\frac{1}{400} \] \[ u = -400 \, \text{cm} \] The negative sign signifies a real object on the incident light's side. The object distance magnitude is \( |u| = 400 \, \text{cm} \). Conversion to meters: \[ |u| = \frac{400}{100} \, \text{m} = 4 \, \text{m} \]