The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real is one-third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is

Options

A. 1m
B. 2m
C. 3m
D. 6m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have a plano-convex lens forming a real image at a distance of 8 m behind the lens. The image size is one-third the size of the object. The wavelength of light inside the lens is \( \frac{2}{3} \) times the wavelength in free space.

### Step 2: Use the magnification formula
The magnification \( m \) is given by:
\[ m = \frac{h'}{h} = \frac{V}{U} \]
Where:
- \( h' \) = height of the image
- \( h \) = height of the object
- \( V \) = image distance
- \( U \) = object distance

Given that the image is one-third the size of the object, we have:
\[ m = -\frac{1}{3} \]
Thus, we can write:
\[ \frac{V}{U} = -\frac{1}{3} \]
This implies:
\[ V = -\frac{1}{3} U \]

### Step 3: Substitute the value of V
We know that \( V = -8 \) m (since the image is real and formed on the opposite side of the object). Therefore:
\[ -8 = -\frac{1}{3} U \]
From this, we can find \( U \):
\[ U = 24 \text{ m} \]

### Step 4: Use the lens formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values of \( V \) and \( U \):
\[ \frac{1}{F} = \frac{1}{-8} - \frac{1}{24} \]
Finding a common denominator (24):
\[ \frac{1}{F} = -\frac{3}{24} - \frac{1}{24} = -\frac{4}{24} = -\frac{1}{6} \]
Thus, we have:
\[ F = -6 \text{ m} \]

### Step 5: Calculate the refractive index
The refractive index \( \mu \) is given by:
\[ \mu = \frac{\lambda_A}{\lambda_M} \]
Where \( \lambda_M = \frac{2}{3} \lambda_A \), thus:
\[ \mu = \frac{\lambda_A}{\frac{2}{3} \lambda_A} = \frac{3}{2} = 1.5 \]

### Step 6: Use the lens maker's formula
The lens maker's formula is:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a plano-convex lens, \( R_1 \) is positive and \( R_2 \) is infinite (flat surface), so:
\[ \frac{1}{F} = (\mu - 1) \frac{1}{R_1} \]
Substituting \( \mu = 1.5 \) and \( F = -6 \):
\[ -\frac{1}{6} = (1.5 - 1) \frac{1}{R_1} \]
This simplifies to:
\[ -\frac{1}{6} = 0.5 \cdot \frac{1}{R_1} \]
Thus:
\[ R_1 = -3 \text{ m} \]

### Step 7: Conclusion
The radius of the curved surface of the lens is \( 3 \) m (the negative sign indicates the direction of curvature).