Question

Light from a point source in air falls on a spherical glass surface (n = 1.5, radius of curvature = 10cm). The distance of the light source from the glass surface is 50 cm. The position at which the image is formed is:

• A. 100 cm in air
• B. 50 cm in glass
• C. 200 cm in air
• D. 150 cm in glass

Answer 1

The Correct Option is B

To solve this problem, we use the lens-maker's formula suitable for spherical surfaces:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} \]
where \( n_1 = 1 \) (refractive index of air), \( n_2 = 1.5 \) (refractive index of glass), \( R = 10 \) cm (radius of curvature), \( u = -50 \) cm (distance of the light source from the surface, negative because it is in the opposite direction to the incident light), and \( v \) is the distance of the image from the surface.

Substitute the known values into the formula:
\[ \frac{1.5}{v} - \frac{1}{-50} = \frac{1.5 - 1}{10} \]
Solving this equation:
\[ \frac{1.5}{v} + \frac{1}{50} = \frac{0.5}{10} \]
\[ \frac{1.5}{v} + \frac{1}{50} = 0.05 \]
\[ \frac{1.5}{v} = 0.05 - \frac{1}{50} \]
\[ \frac{1.5}{v} = 0.05 - 0.02 \]
\[ \frac{1.5}{v} = 0.03 \]
\[ v = \frac{1.5}{0.03} \]
\[ v = 50 \] cm

Thus, the image is formed 50 cm inside the glass.

Answer 2

• Point source in air (n₁ = 1.0)
• Spherical glass surface (n₂ = 1.5)
• Radius of curvature (R) = 10 cm
• Object distance (u) = -50 cm (negative sign convention for real object)

Using the Refraction at Spherical Surface Formula:

\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

Substituting Values:

\[ \frac{1.5}{v} - \frac{1.0}{-50} = \frac{1.5 - 1.0}{10} \]

\[ \frac{1.5}{v} + \frac{1}{50} = \frac{0.5}{10} \]

\[ \frac{1.5}{v} = 0.05 - 0.02 = 0.03 \]

\[ v = \frac{1.5}{0.03} = 50 \text{ cm} \]

Interpretation:

• Positive value of v indicates the image is formed on the opposite side of the glass (real image)
• Image distance = 50 cm from the spherical surface