Distance of an object from the first focus of an equiconvex lens is 10 cm and the distance of its reimage from second focus is 40 cm. The focal length the lens is

Options

A. 25 cm
B. 10 cm
C. 20 cm
D. 40 cm

Answer

To solve the problem, we will use the lens formula, which is given by:

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

Where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 1: Identify the given values
From the problem:
- The distance of the object from the first focus \( F_1 \) is given as \( 10 \, \text{cm} \). Since the object is placed at the first focus, we can say that the object distance \( u = -10 \, \text{cm} \) (the negative sign is used as per the sign convention).
- The distance of the image from the second focus \( F_2 \) is given as \( 40 \, \text{cm} \). Therefore, the image distance \( v = +40 \, \text{cm} \) (the positive sign indicates that the image is on the opposite side of the lens).

### Step 2: Substitute the values into the lens formula
Now, we can substitute the values of \( u \) and \( v \) into the lens formula:

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

Substituting the values:

\[ \frac{1}{f} = \frac{1}{40} - \frac{1}{-10} \]

### Step 3: Simplify the equation
Calculating the right side:

\[ \frac{1}{f} = \frac{1}{40} + \frac{1}{10} \]

To add these fractions, we need a common denominator. The least common multiple of 40 and 10 is 40:

\[ \frac{1}{f} = \frac{1}{40} + \frac{4}{40} = \frac{5}{40} \]

### Step 4: Solve for \( f \)
Now, we can solve for \( f \):

\[ \frac{1}{f} = \frac{5}{40} \]

Taking the reciprocal:

\[ f = \frac{40}{5} = 8 \, \text{cm} \]

### Step 5: Verify the focal length
Since the problem states that the lens is equiconvex, both focal lengths are equal. Therefore, the focal length of the lens is:

\[ f = 8 \, \text{cm} \]

### Final Answer
The focal length of the lens is \( 8 \, \text{cm} \).

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