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

Options

A. 1m
B. 2m
C. 3m
D. 4m

Explanation

To solve the problem step-by-step, we will follow the information provided in the question and apply the relevant formulas in geometrical optics.

### Step 1: Understanding the Given Data
We have a plano-convex lens with the following information:
- The image distance (v) is 8 m (real image, so v is positive).
- The image size is one-third the size of the object (height of image \( h_i = \frac{1}{3} h_o \)).
- The wavelength of light inside the lens is \( \frac{2}{3} \) times the wavelength in free space.

### Step 2: Using the Magnification Formula
The magnification (m) for a lens is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Where:
- \( h_i \) is the height of the image,
- \( h_o \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.

Since \( h_i = \frac{1}{3} h_o \), we can write:
\[ m = \frac{1}{3} \]
Thus:
\[ \frac{v}{u} = \frac{1}{3} \]
Substituting \( v = 8 \) m:
\[ \frac{8}{u} = \frac{1}{3} \]
Cross-multiplying gives:
\[ u = 8 \times 3 = 24 \text{ m} \]
Since the object is on the opposite side of the lens from the image, we take \( u = -24 \) m.

### Step 3: Using the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{8} - \frac{1}{-24} \]
Calculating the right-hand side:
\[ \frac{1}{f} = \frac{1}{8} + \frac{1}{24} \]
Finding a common denominator (24):
\[ \frac{1}{f} = \frac{3}{24} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6} \]
Thus, the focal length \( f \) is:
\[ f = 6 \text{ m} \]

### Step 4: Finding the Refractive Index
Given that the wavelength of light inside the lens is \( \frac{2}{3} \) times the wavelength in free space, we can find the refractive index \( \mu \):
\[ \mu = \frac{\text{Wavelength in air}}{\text{Wavelength in lens}} = \frac{\lambda}{\frac{2}{3} \lambda} = \frac{3}{2} \]

### Step 5: Applying the Lens Maker's Formula
The lens maker's formula is:
\[ \frac{1}{f} = \mu - 1 \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a plano-convex lens:
- \( R_1 = r \) (the radius of curvature of the convex surface),
- \( R_2 = \infty \) (the plane surface).

Thus, we have:
\[ \frac{1}{f} = \left(\frac{3}{2} - 1\right) \cdot \left(\frac{1}{r} - 0\right) \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{2} \cdot \frac{1}{r} \]
Substituting \( f = 6 \):
\[ \frac{1}{6} = \frac{1}{2} \cdot \frac{1}{r} \]
Cross-multiplying gives:
\[ r = 3 \text{ m} \]

### Final Answer
The radius of the curved surface of the lens is \( r = 3 \text{ m} \).

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