A convex lens of focal length 30 cm forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the shift of the object is 6 cm. The shift of screen is

Options

A. 28 cm
B. 14 cm
C. 18 cm
D. 16 cm

Explanation

To solve the problem step by step, we will follow the principles of optics, specifically using the lens formula and magnification concepts.

### Step 1: Understand the Given Information
We have a convex lens with a focal length (f) of 30 cm. The initial magnification (M1) is 3, which means the image formed is three times larger than the object.

### Step 2: Use the Magnification Formula
The magnification (M) is given by the formula:
\[ M = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance.

From the problem, we know:
\[ M_1 = 3 = \frac{v_1}{u_1} \]
This implies:
\[ v_1 = 3u_1 \]

### Step 3: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 30 \) cm, \( v = v_1 \), and \( u = u_1 \):
\[ \frac{1}{30} = \frac{1}{v_1} - \frac{1}{u_1} \]

Substituting \( v_1 = 3u_1 \):
\[ \frac{1}{30} = \frac{1}{3u_1} - \frac{1}{u_1} \]

### Step 4: Solve for \( u_1 \)
Rearranging the equation:
\[ \frac{1}{30} = \frac{1 - 3}{3u_1} = \frac{-2}{3u_1} \]
This leads to:
\[ 3u_1 = -60 \]
Thus:
\[ u_1 = -20 \text{ cm} \]

### Step 5: Calculate \( v_1 \)
Now substituting \( u_1 \) back to find \( v_1 \):
\[ v_1 = 3(-20) = -60 \text{ cm} \]

### Step 6: New Magnification Condition
Now the object and screen are moved until the image becomes twice the size of the object. Thus, the new magnification \( M_2 = 2 \):
\[ M_2 = \frac{v_2}{u_2} \]
This implies:
\[ v_2 = 2u_2 \]

### Step 7: Object Shift
The object is shifted 6 cm. Therefore, if the original object distance was \( u_1 = -20 \) cm, the new object distance is:
\[ u_2 = -20 + 6 = -14 \text{ cm} \]

### Step 8: Apply the Lens Formula Again
Using the lens formula again:
\[ \frac{1}{30} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( u_2 = -14 \):
\[ \frac{1}{30} = \frac{1}{v_2} + \frac{1}{14} \]

### Step 9: Solve for \( v_2 \)
Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{30} - \frac{1}{14} \]
Finding a common denominator (LCM of 30 and 14 is 210):
\[ \frac{1}{v_2} = \frac{7 - 15}{210} = \frac{-8}{210} \]
Thus:
\[ v_2 = -\frac{210}{8} = -26.25 \text{ cm} \]

### Step 10: Calculate the Shift of the Screen
The shift of the screen is given by:
\[ \text{Shift of screen} = v_1 - v_2 \]
Substituting the values:
\[ \text{Shift of screen} = -60 - (-26.25) = -60 + 26.25 = -33.75 \text{ cm} \]

### Conclusion
The shift of the screen is approximately 33.75 cm to the left.