The image of a square hole in a screen illuminated by light is obtained on another screen with the help of converging lens. The distance of the hole from the lens is 40 cm. If the area of the image is nine times that of the hole, the focal length of the lens is

Options

A. 30cm
B. 50cm
C. 60cm
D. 75cm

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the given data
We have a square hole in a screen that is illuminated and its image is formed on another screen using a converging lens (convex lens). The distance of the hole from the lens (u) is given as 40 cm. The area of the image is nine times that of the hole.

### Step 2: Calculate the linear magnification
The area magnification (M_area) is given as 9. The relationship between area magnification and linear magnification (M_linear) is:
\[ M_{linear} = \sqrt{M_{area}} \]
Thus,
\[ M_{linear} = \sqrt{9} = 3 \]

### Step 3: Relate magnification to object and image distances
The magnification (M) in terms of object distance (u) and image distance (v) is given by:
\[ M = -\frac{v}{u} \]
Since we have a linear magnification of 3, we can write:
\[ 3 = -\frac{v}{u} \]
Given that \( u = -40 \) cm (the negative sign indicates that the object is on the same side as the incoming light), we can substitute:
\[ 3 = -\frac{v}{-40} \]
This simplifies to:
\[ v = 3 \times 40 = 120 \text{ cm} \]

### Step 4: Use the lens formula to find the focal length
The lens formula relates the object distance (u), image distance (v), and focal length (f):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{-40} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{120} + \frac{1}{40} \]

### Step 5: Find a common denominator and calculate
The common denominator for 120 and 40 is 120. Thus, we can rewrite:
\[ \frac{1}{40} = \frac{3}{120} \]
Now substituting back:
\[ \frac{1}{f} = \frac{1}{120} + \frac{3}{120} = \frac{4}{120} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{30} \]
Thus, the focal length \( f \) is:
\[ f = 30 \text{ cm} \]

### Final Answer
The focal length of the lens is **30 cm**.

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