The distance between the object and its real image from the convex lens is 60 cm and the height of image is two times the height of object . The focal length of the lens is

Options

A. `20/3` cm
B. 20 cm
C. `40/3 cm`
D. 40 cm

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula.

### Step 1: Understand the Given Information
- The distance between the object (O) and its real image (I) is given as 60 cm.
- The height of the image (I) is twice the height of the object (O), which means the magnification (m) is 2.

### Step 2: Set Up the Variables
Let:
- \( u \) = distance of the object from the lens (in cm)
- \( v \) = distance of the image from the lens (in cm)

From the problem, we know:
\[ v - u = 60 \, \text{cm} \] (1)

### Step 3: Use the Magnification Formula
The magnification (m) is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Since \( m = 2 \):
\[ \frac{v}{u} = 2 \]
This implies:
\[ v = 2u \] (2)

### Step 4: Substitute Equation (2) into Equation (1)
Substituting \( v = 2u \) into \( v - u = 60 \):
\[ 2u - u = 60 \]
\[ u = 60 \, \text{cm} \]

### Step 5: Find the Distance of the Image
Now, substituting \( u \) back into equation (2):
\[ v = 2u = 2 \times 60 = 120 \, \text{cm} \]

### Step 6: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{-60} \]
\[ \frac{1}{f} = \frac{1}{120} + \frac{1}{60} \]

### Step 7: Simplify the Right Side
Finding a common denominator (which is 120):
\[ \frac{1}{f} = \frac{1}{120} + \frac{2}{120} = \frac{3}{120} \]
Thus:
\[ \frac{1}{f} = \frac{1}{40} \]

### Step 8: Solve for the Focal Length
Taking the reciprocal gives:
\[ f = 40 \, \text{cm} \]

### Final Answer
The focal length of the lens is \( 40 \, \text{cm} \).

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