When an object is placed 40cm from a diverging lens, its virtual image is formed 20cm from the lens. The focal length and power of lens are

Options

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B.
C.
D.

Explanation

To solve the problem step by step, we will use the lens formula and the definition of power of a lens.

### Step 1: Identify the given values
- Object distance (u) = -40 cm (negative because the object is on the same side as the incoming light for a diverging lens)
- Image distance (v) = -20 cm (negative because the virtual image is formed on the same side as the object)

### Step 2: Use the lens formula
The lens formula 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 3: Substitute the values into the lens formula
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{-20} - \frac{1}{-40} \]

### Step 4: Simplify the equation
Calculating the right-hand side:
\[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{40} \]
To combine the fractions, find a common denominator (which is 40):
\[ \frac{1}{f} = -\frac{2}{40} + \frac{1}{40} = -\frac{2 - 1}{40} = -\frac{1}{40} \]

### Step 5: Solve for the focal length (f)
Taking the reciprocal gives:
\[ f = -40 \text{ cm} \]

### Step 6: Calculate the power of the lens
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
First, convert the focal length from centimeters to meters:
\[ f = -40 \text{ cm} = -0.4 \text{ m} \]
Now, substituting into the power formula:
\[ P = \frac{1}{-0.4} = -2.5 \text{ diopters} \]

### Final Answers
- Focal length \( f = -40 \text{ cm} \)
- Power \( P = -2.5 \text{ diopters} \)