In displacement method, the distance between object and screen is 96cm. The ratio of lengths of two images formed by a converging lens placed between them is 4. Then,

Options

A. ratio of the length of object to the length of shorter image is 2
B. distance between the two positions of the lens is 32 cm
C. focal length of the lens is `64//3` cm
D. when the shorter image is formed on screen, distance of the lens from the screen is 32cm.

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant concepts of optics.

### Step 1: Understand the given data
- The total distance between the object and the screen is \(96 \, \text{cm}\).
- The ratio of the lengths of two images formed by the converging lens is \(4:1\).
- The distance between the two positions of the lens is \(32 \, \text{cm}\).
- We need to find the focal length of the lens and other relevant parameters.

### Step 2: Set up the equations
Let:
- \(u_1\) = object distance for the first position of the lens
- \(v_1\) = image distance for the first position of the lens
- \(u_2\) = object distance for the second position of the lens
- \(v_2\) = image distance for the second position of the lens

From the problem, we know:
\[ u_1 + v_1 = 96 \quad \text{(1)} \]
\[ u_2 + v_2 = 96 \quad \text{(2)} \]

### Step 3: Relate the image heights to the object height
Let the height of the object be \(O\), the height of the first image be \(I_1\), and the height of the second image be \(I_2\). The magnification for each lens position can be expressed as:
\[ \frac{I_1}{O} = \frac{v_1}{u_1} \quad \text{(3)} \]
\[ \frac{I_2}{O} = \frac{u_2}{v_2} \quad \text{(4)} \]

### Step 4: Use the ratio of the image heights
Given that the ratio of the lengths of the two images is \(4:1\), we can express this as:
\[ \frac{I_1}{I_2} = 4 \quad \Rightarrow \quad I_1 = 4I_2 \]

### Step 5: Substitute and solve for \(u\) and \(v\)
From equations (3) and (4), we can express \(I_1\) and \(I_2\) in terms of \(u\) and \(v\):
\[ \frac{I_1}{I_2} = \frac{v_1/u_1}{u_2/v_2} = 4 \]

Substituting \(u_2 = 96 - v_1\) into the magnification ratio gives:
\[ \frac{v_1}{u_1} \cdot \frac{v_2}{96 - v_1} = 4 \]

### Step 6: Solve for distances
From the ratio, we can derive:
\[ v_1 = 4 \cdot \frac{u_1}{v_2} \cdot (96 - v_1) \]
Using the distance between the two lens positions:
\[ u_2 = u_1 + 32 \]

### Step 7: Solve for \(u_1\) and \(v_1\)
From the equations, we can solve for \(u_1\) and \(v_1\). After substituting and simplifying, we find:
\[ u_1 = 32 \, \text{cm}, \quad v_1 = 64 \, \text{cm} \]

### Step 8: Calculate the focal length
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \(v = 64 \, \text{cm}\) and \(u = -32 \, \text{cm}\) (using sign conventions):
\[ \frac{1}{f} = \frac{1}{64} + \frac{1}{32} \]
Calculating gives:
\[ \frac{1}{f} = \frac{3}{64} \quad \Rightarrow \quad f = \frac{64}{3} \, \text{cm} \]

### Final Results
- The object distance \(u = 32 \, \text{cm}\)
- The image distance \(v = 64 \, \text{cm}\)
- The focal length \(f = \frac{64}{3} \, \text{cm}\)