A point object is placed at a distance 25cm from a convex lens of focal length 20cm. If a glass slab of thickness t annd refractive index 1.5 is inserted between the lens and object, image is formed at . Thickness t is found to be K times of 5cm. Fink K.

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript while elaborating on the calculations involved.

### Step 1: Understand the Setup
We have a point object placed at a distance of 25 cm from a convex lens with a focal length of 20 cm. When a glass slab of thickness \( t \) and refractive index \( \mu = 1.5 \) is inserted between the lens and the object, the image is formed at infinity.

### Step 2: Determine the Condition for Image Formation at Infinity
For the image to be formed at infinity, the object must be at the focal point of the lens. The focal length \( f \) of the lens is 20 cm, so the effective object distance \( u \) must be equal to the focal length when the glass slab is inserted.

### Step 3: Calculate the Shift Due to the Glass Slab
The original object distance is 25 cm. When the glass slab is inserted, the object effectively shifts towards the lens. The shift \( S \) caused by the slab can be calculated using the formula:

\[ S = t \left(1 - \frac{1}{\mu}\right) \]

where \( t \) is the thickness of the slab and \( \mu \) is the refractive index of the slab.

### Step 4: Substitute Values into the Shift Formula
Given \( \mu = 1.5 \):

\[ S = t \left(1 - \frac{1}{1.5}\right) = t \left(1 - \frac{2}{3}\right) = t \left(\frac{1}{3}\right) \]

### Step 5: Relate the Shift to the Object Distance
Since the object is originally at 25 cm and must shift to the focal length of 20 cm, the shift \( S \) is:

\[ S = 25 \text{ cm} - 20 \text{ cm} = 5 \text{ cm} \]

### Step 6: Set Up the Equation
Now we can set up the equation using the shift we calculated:

\[ 5 = t \left(\frac{1}{3}\right) \]

### Step 7: Solve for Thickness \( t \)
Rearranging the equation gives:

\[ t = 5 \times 3 = 15 \text{ cm} \]

### Step 8: Relate Thickness \( t \) to \( K \)
The problem states that the thickness \( t \) is found to be \( K \) times of 5 cm:

\[ t = K \times 5 \text{ cm} \]

Substituting \( t = 15 \text{ cm} \):

\[ 15 = K \times 5 \]

### Step 9: Solve for \( K \)
Dividing both sides by 5:

\[ K = \frac{15}{5} = 3 \]

### Final Answer
Thus, the value of \( K \) is 3.

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