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 and its real image from the convex lens is 60 cm.
- The height of the image (I) is twice the height of the object (O), so \( I = 2O \).

### Step 2: Define 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:
\[ u + v = 60 \, \text{cm} \]

### Step 3: Use the magnification formula
The magnification \( M \) is given by:
\[ M = \frac{I}{O} = \frac{v}{u} \]
Since \( I = 2O \), we have:
\[ M = 2 \]
Thus:
\[ \frac{v}{u} = 2 \]
This implies:
\[ v = 2u \]

### Step 4: Substitute \( v \) in the distance equation
Substituting \( v = 2u \) into the equation \( u + v = 60 \):
\[ u + 2u = 60 \]
\[ 3u = 60 \]
\[ u = 20 \, \text{cm} \]

### Step 5: Find \( v \)
Now, substituting \( u = 20 \) cm back into the equation for \( v \):
\[ v = 2u = 2 \times 20 = 40 \, \text{cm} \]

### 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 \( u \):
\[ \frac{1}{f} = \frac{1}{40} - \frac{1}{20} \]

### Step 7: Calculate \( \frac{1}{f} \)
Finding a common denominator (which is 40):
\[ \frac{1}{f} = \frac{1}{40} - \frac{2}{40} = -\frac{1}{40} \]

### Step 8: Find \( f \)
Taking the reciprocal gives:
\[ f = -40 \, \text{cm} \]
Since we are dealing with a convex lens, we take the absolute value:
\[ f = 40 \, \text{cm} \]

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

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