A point object is placed at a diatance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness t is

Options

A. (a)10 cm
B. (b)5 cm
C. (c)20 cm
D. (d)15 cm

Answer

To solve the problem step by step, we will use the lens formula and the concept of refraction through a glass slab.

### Step 1: Understand the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 2: Identify the given values
From the problem:
- Focal length of the lens, \( f = 20 \) cm
- Object distance without the slab, \( u = -25 \) cm (the negative sign indicates that the object is on the same side as the incoming light)

### Step 3: Calculate the image distance without the slab
Using the lens formula:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{(-25)} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{20} + \frac{1}{25} \]
Finding a common denominator (100):
\[ \frac{1}{v} = \frac{5}{100} + \frac{4}{100} = \frac{9}{100} \]
Thus,
\[ v = \frac{100}{9} \approx 11.11 \text{ cm} \]
This means that the image is formed at approximately 11.11 cm on the opposite side of the lens.

### Step 4: Understand the effect of the glass slab
When a glass slab of thickness \( t \) and refractive index \( \mu = 1.5 \) is inserted, the effective object distance changes. The image is said to be formed at infinity, which means the object must be at the focal point of the lens.

### Step 5: Calculate the new object distance with the slab
For the image to be formed at infinity, the object distance \( u' \) must equal the focal length \( f \):
\[ u' = -f = -20 \text{ cm} \]
The new object distance \( u' \) can be expressed as:
\[ u' = u - t \left(1 - \frac{1}{\mu}\right) \]
Substituting the values:
\[ -20 = -25 - t \left(1 - \frac{1}{1.5}\right) \]
Calculating \( 1 - \frac{1}{1.5} \):
\[ 1 - \frac{1}{1.5} = 1 - \frac{2}{3} = \frac{1}{3} \]
So, we have:
\[ -20 = -25 - t \left(\frac{1}{3}\right) \]
Rearranging gives:
\[ -20 + 25 = -\frac{t}{3} \]
\[ 5 = -\frac{t}{3} \]
Multiplying both sides by -3:
\[ t = -15 \text{ cm} \]
Since thickness cannot be negative, we take the absolute value:
\[ t = 15 \text{ cm} \]

### Final Answer
The thickness \( t \) of the glass slab is \( 15 \) cm.

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