Question:

Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at____ cm from the object.

Updated On: Nov 5, 2025
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Correct Answer: 200

Approach Solution - 1

Given:

\( \mu_1 = 1, \quad \mu_2 = 1.5, \quad R = 20 \, \text{cm}, \quad \text{Object distance} = 100 \, \text{cm} \)

Step 1: Using the refraction formula

The relation between the refractive indices and distances is given by: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

Step 2: Substituting the known values

\[ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20} \]

Step 3: Solving for \( v \)

\[ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} \]

\[ \frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100} \]

\[ \frac{1.5}{v} = \frac{5}{100} - \frac{1}{100} = \frac{4}{100} \]

\[ v = \frac{1.5 \times 100}{4} = 100 \, \text{cm} \]

Step 4: Finding the total distance

The total distance from the object is: \[ \text{Distance from object} = 100 + 100 = 200 \, \text{cm} \]

Final Answer:

\[ \boxed{200 \, \text{cm}} \]

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Approach Solution -2

Given: - Radius of curvature \( R = 20 \, \text{cm} \), - Refractive indices: \( \mu_1 = 1 \) (air) and \( \mu_2 = 1.5 \) (medium).

Using the lens maker’s formula:

\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. \]

Substitute \( u = -100 \, \text{cm} \):

\[ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}. \]

Solving for \( v \):

\[ \frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}, \] \[ \frac{1.5}{v} = \frac{0.5}{20} - \frac{1}{100}, \] \[ v = 100 \, \text{cm}. \]

Thus, the image is formed at a distance:

\[ 100 + 100 = 200 \, \text{cm from the object}. \]
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