A ray of light is incident normally on a glass slab of thickness 5 cm and refractive index 1.6. The time taken to travel by a ray from source to surface of slab is same as to travel through glass slab. The distance of source from the surface is

Options

A. 4 cm
B. 8 cm
C. 12 cm
D. 16 cm

Answer

To solve the problem step by step, we will analyze the situation using the information provided.

### Step 1: Understand the Problem
We have a glass slab of thickness \( t = 5 \, \text{cm} \) and a refractive index \( n = 1.6 \). A ray of light is incident normally on the slab, and the time taken to travel from the source to the surface of the slab is the same as the time taken to travel through the glass slab.

### Step 2: Define Variables
Let:
- \( S_1 \) = distance from the source to the surface of the slab
- \( S_2 \) = distance traveled through the glass slab (which is equal to the thickness of the slab, \( t = 5 \, \text{cm} \))
- \( c \) = speed of light in vacuum
- \( v \) = speed of light in the glass slab

### Step 3: Calculate Speed in Glass
The speed of light in the glass slab can be calculated using the formula:
\[ v = \frac{c}{n} \]
Substituting the value of \( n \):
\[ v = \frac{c}{1.6} \]

### Step 4: Set Up Time Equations
Since the time taken to travel from the source to the surface of the slab is equal to the time taken to travel through the glass slab, we can write:
\[ t_1 = t_2 \]
Where:
\[ t_1 = \frac{S_1}{c} \quad \text{(time taken to travel from source to surface)} \]
\[ t_2 = \frac{S_2}{v} \quad \text{(time taken to travel through the glass slab)} \]

### Step 5: Substitute Values
Substituting \( S_2 = 5 \, \text{cm} \) and \( v = \frac{c}{1.6} \) into the time equation:
\[ \frac{S_1}{c} = \frac{5}{\frac{c}{1.6}} \]
This simplifies to:
\[ \frac{S_1}{c} = \frac{5 \cdot 1.6}{c} \]

### Step 6: Eliminate \( c \) and Solve for \( S_1 \)
Since \( c \) is present in both sides, we can eliminate it:
\[ S_1 = 5 \cdot 1.6 \]
Calculating this gives:
\[ S_1 = 8 \, \text{cm} \]

### Step 7: Conclusion
The distance of the source from the surface of the slab is \( 8 \, \text{cm} \).

### Final Answer
The distance of the source from the surface of the slab is \( 8 \, \text{cm} \).

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