A plano convex lens of refractive index 1.5 and radius of curvature 30cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object.

Options

A. 60 cm
B. 30 cm
C. 20 cm
D. 80 cm

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understanding the system
We have a plano-convex lens that is silvered on its curved surface. When the lens is silvered, it behaves like a mirror for light that is reflected from the silvered surface. The lens will refract light as it passes through, and then the silvered surface will reflect it back.

### Step 2: Finding the focal length of the lens
The focal length (f) of a plano-convex lens can be calculated using the lens maker's formula:

\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For a plano-convex lens:
- \( n = 1.5 \) (refractive index)
- \( R_1 = +30 \, \text{cm} \) (radius of curvature of the convex surface)
- \( R_2 = \infty \) (the flat surface)

Substituting the values:

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{30} - 0 \right) = 0.5 \cdot \frac{1}{30} = \frac{1}{60} \]

Thus, the focal length \( f \) is:

\[ f = 60 \, \text{cm} \]

### Step 3: Considering the silvered surface
When the lens is silvered, the effective focal length of the system changes. The silvered surface acts like a concave mirror. The focal length of a concave mirror is negative, so we can use the formula for the effective focal length \( f' \) of the lens-mirror combination:

\[ \frac{1}{f'} = \frac{1}{f} + \frac{1}{f_m} \]

Where \( f_m = -30 \, \text{cm} \) (focal length of the mirror).

Substituting the values:

\[ \frac{1}{f'} = \frac{1}{60} - \frac{1}{30} \]

Calculating the right-hand side:

\[ \frac{1}{f'} = \frac{1}{60} - \frac{2}{60} = -\frac{1}{60} \]

Thus, the effective focal length \( f' \) is:

\[ f' = -60 \, \text{cm} \]

### Step 4: Using the lens formula
Now we can use the lens formula to find the object distance \( u \) for a real image of the same size as the object:

\[ \frac{1}{f'} = \frac{1}{v} - \frac{1}{u} \]

For a real image of the same size, the image distance \( v \) is equal to the object distance \( u \). Thus, we can set \( v = -u \):

\[ \frac{1}{-60} = \frac{1}{-u} - \frac{1}{u} \]

This simplifies to:

\[ \frac{1}{-60} = -\frac{2}{u} \]

Cross-multiplying gives:

\[ -2 = -\frac{u}{60} \]

Thus:

\[ u = 120 \, \text{cm} \]

### Final Answer
The object should be placed at a distance of **120 cm** from the lens to obtain a real image of the same size as the object.

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