Question

In a long glass tube, a mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of \(-2\), then the radius of curvature of the meniscus is:

• A. \( \frac{1}{3} \) cm
• B. 1 cm
• C. \( \frac{4}{3} \) cm
• D. \( \frac{2}{3} \) cm

Hint:

For refraction problems, always use sign conventions properly to avoid errors.

Answer 1

The Correct Option is D

• Using the lens-maker's formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
• Substituting given values: \[ \frac{1.4}{v} - \frac{1.3}{-13} = \frac{0.1}{R} \]
• Simplifying the equation: \[ \frac{1.4}{v} = \frac{1 - R}{10R} \]
• From magnification: \[ m = \frac{v/n_2}{u/n_1} \] \[ -2 \times (-13)/1.3 = 10R / (1 - R) \]
• Finally: \[ R = \frac{2}{3} \, \text{cm} \]

Answer 2

Step 1: Given data.
Refractive index of medium A = \( \mu_1 = 1.3 \)
Refractive index of medium B = \( \mu_2 = 1.4 \)
Object distance \( u = -13 \, \text{cm} \) (object in medium A, measured from the vertex of the meniscus)
Magnification \( m = -2 \)
We need to find the radius of curvature \( R \) of the convex refracting surface (convex towards A).

Step 2: Relation between magnification and image distance.
For refraction at a spherical surface, linear magnification is given by:
\[ m = \frac{\mu_1 u}{\mu_2 v} \] Substitute the given values:
\[ -2 = \frac{1.3 \times (-13)}{1.4 \times v} \] \[ -2 = \frac{16.9}{1.4v} \] \[ v = \frac{16.9}{2.8} = 6.04 \, \text{cm}. \] So, \( v = +6.04 \, \text{cm} \).

Step 3: Apply the refraction formula at a spherical surface.
The refraction formula between two media is:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] Substitute values:
\[ \frac{1.4}{6.04} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R} \] \[ 0.2317 + 0.1 = \frac{0.1}{R} \] \[ 0.3317 = \frac{0.1}{R} \] \[ R = \frac{0.1}{0.3317} = 0.301 \, \text{cm} \approx \frac{2}{3} \, \text{cm}. \]

Step 4: Final Answer.
The radius of curvature of the meniscus is:
\[ \boxed{R = \frac{2}{3} \, \text{cm}} \]