An object is placed 30cm away from a convex lens of focal length 10cm and a sharp image is formed on a screen. Now, a convex lens is placed In contact with the convex lens. The screen now has to be moved by 45cm to get a sharp image again. The magnitude of focal length of the convex lends in (in cm)

Options

A. 72
B. 60
C. 36
D. 20

Explanation

To solve the problem step-by-step, we will use the lens formula and the information provided in the question.

### Step 1: Identify the given parameters
- Object distance from the first lens, \( u_1 = -30 \, \text{cm} \) (negative as per sign convention)
- Focal length of the first lens, \( f_1 = 10 \, \text{cm} \)
- The distance the screen is moved to form a sharp image again, \( d = 45 \, \text{cm} \)

### Step 2: Use the lens formula to find the image distance for the first lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the first lens, we can rearrange it to find the image distance \( v_1 \):
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \]
Substituting the known values:
\[ \frac{1}{10} = \frac{1}{v_1} - \frac{1}{(-30)} \]
\[ \frac{1}{10} = \frac{1}{v_1} + \frac{1}{30} \]
Now, we find a common denominator (30):
\[ \frac{1}{10} = \frac{3}{30} \quad \text{and} \quad \frac{1}{30} = \frac{1}{30} \]
Thus:
\[ \frac{3}{30} = \frac{1}{v_1} + \frac{1}{30} \]
Subtracting \( \frac{1}{30} \) from both sides:
\[ \frac{3}{30} - \frac{1}{30} = \frac{1}{v_1} \]
\[ \frac{2}{30} = \frac{1}{v_1} \]
Inverting gives:
\[ v_1 = \frac{30}{2} = 15 \, \text{cm} \]

### Step 3: Determine the total image distance after moving the screen
The total distance from the first lens to the new image position after moving the screen is:
\[ v_2 = v_1 + d = 15 \, \text{cm} + 45 \, \text{cm} = 60 \, \text{cm} \]

### Step 4: Use the lens formula for the combination of lenses
Now, we need to find the focal length of the second lens, \( f_2 \), when both lenses are in contact. The object distance remains the same, \( u_2 = -30 \, \text{cm} \), and we can use the lens formula again:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \]
Where \( f \) is the equivalent focal length of the two lenses combined, which we denote as \( f \):
\[ \frac{1}{f} = \frac{1}{60} - \frac{1}{(-30)} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{60} + \frac{1}{30} \]
Finding a common denominator (60):
\[ \frac{1}{30} = \frac{2}{60} \]
Thus:
\[ \frac{1}{f} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} \]
So:
\[ f = \frac{60}{3} = 20 \, \text{cm} \]

### Step 5: Find the focal length of the second lens
Using the formula for the combination of two lenses:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
Substituting the known values:
\[ \frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \]
Rearranging gives:
\[ \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} \]
Finding a common denominator (20):
\[ \frac{1}{10} = \frac{2}{20} \]
Thus:
\[ \frac{1}{f_2} = \frac{1}{20} - \frac{2}{20} = -\frac{1}{20} \]
So:
\[ f_2 = -20 \, \text{cm} \]

### Step 6: Find the magnitude of the focal length
Since we are asked for the magnitude:
\[ |f_2| = 20 \, \text{cm} \]

### Final Answer
The magnitude of the focal length of the convex lens in cm is \( 20 \, \text{cm} \).

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