A convex lens forms an image of an object on a screen. The height of the image is 9 cm . The lens is now displaced until an image is again obtained on the screen. Then height of this image is 4 cm . The distance between the object and the screen is 90 cm.

Options

A. The distance between the two positions of the lens is 30 cm
B. The distance of the object from the lens in its first position is 36 cm
C. The height of the object is 6 cm
D. The focal length of the lens is 21.6 cm

Explanation

To solve the problem step by step, we will use the information provided about the heights of the images formed by the convex lens and the distance between the object and the screen.

### Step 1: Understanding the Given Information
- Height of the first image (h1) = 9 cm
- Height of the second image (h2) = 4 cm
- Distance between the object and the screen (D) = 90 cm

### Step 2: Finding the Height of the Object
We can use the relationship between the heights of the images and the height of the object. The formula for the height of the object (h) in terms of the heights of the images is given by:

\[ h = \sqrt{h_1 \times h_2} \]

Substituting the values:

\[ h = \sqrt{9 \times 4} = \sqrt{36} = 6 \text{ cm} \]

### Step 3: Finding the Linear Magnification
The linear magnification (m) is given by:

\[ m = \frac{h}{h_o} = \frac{h_1}{h_o} = \frac{V_1}{U_1} \]

Where:
- \(h_o\) is the height of the object,
- \(V_1\) is the image distance for the first image,
- \(U_1\) is the object distance for the first image.

Using the magnification for the first image:

\[ m_1 = \frac{9}{6} = \frac{3}{2} \]

### Step 4: Finding the Object Distance for the Second Image
Using the magnification for the second image:

\[ m_2 = \frac{h_2}{h_o} = \frac{4}{6} = \frac{2}{3} \]

We can also express the relationship between object distance and image distance as follows:

\[ U_2 = \frac{2}{3} U_1 \]

### Step 5: Setting Up the Equation for Total Distance
The total distance between the object and the screen is given as:

\[ D = U_1 + U_2 \]

Substituting \(U_2\):

\[ 90 = U_1 + \frac{2}{3} U_1 \]

This simplifies to:

\[ 90 = \frac{5}{3} U_1 \]

### Step 6: Solving for \(U_1\)
To find \(U_1\):

\[ U_1 = 90 \times \frac{3}{5} = 54 \text{ cm} \]

### Step 7: Finding \(U_2\)
Now, substituting \(U_1\) back to find \(U_2\):

\[ U_2 = \frac{2}{3} \times 54 = 36 \text{ cm} \]

### Step 8: Finding the Focal Length
Using the lens formula:

\[ \frac{1}{f} = \frac{1}{V} - \frac{1}{U} \]

We can find the focal length \(f\) using the distances:

\[ D = U_1 + U_2 = 54 + 36 = 90 \text{ cm} \]

Using the distances in the lens formula, we can find the focal length.

### Final Calculation for Focal Length
Using the derived values, we can calculate the focal length \(f\) based on the distances.

### Conclusion
The height of the object is 6 cm, and the object distances for the two images are \(U_1 = 54 \text{ cm}\) and \(U_2 = 36 \text{ cm}\).

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