Question

For the given YDSE setup. Find the number of fringes by which the central maxima gets shifted from point O.
(Given d = 1 mm D = 1 m λ = 5000 Å)

• A. 10
• B. 15
• C. 8
• D. 12

Answer 1

The Correct Option is A

At the central position, the path difference is given by: 

\[ \Delta x = (\mu - 1)t_1 - (\mu - 1)t_2 \]

Factoring \((\mu - 1)\), we get:

\[ \Delta x = (\mu - 1)(t_1 - t_2) \]

Substitute the given values:

\[ \Delta x = \left(\frac{3}{2} - 1\right)(5.11 - 5.10) \ \text{mm} \]

\[ \Delta x = \frac{1}{2} \times 0.001 \ \text{mm} \]

\[ \Delta x = 0.0005 \ \text{mm} = 5 \times 10^{-6} \ \text{m} \]

The number of fringes shifted is given by:

\[ \text{No. of fringes shifted} = \frac{\Delta x}{\lambda} \]

Substitute the values \(\Delta x = 5 \times 10^{-6} \ \text{m}\) and \(\lambda = 5 \times 10^{-7} \ \text{m}\):

\[ \text{No. of fringes shifted} = \frac{5 \times 10^{-6}}{5 \times 10^{-7}} \]

\[ \text{No. of fringes shifted} = 10 \]

Final Answer:

(A) : 10