Question

In Young's double slit experiment, the slit width is \(1\) mm and screen is at a distance \(2\) m from slits. Light of \(600\) nm is used. If a thin glass plate of refractive index \(1.5\) of thickness \(0.06\) mm is placed over one of the slits then there will be a lateral displacement of the fringes by

• A. \(0\) cm
• B. \(6\) cm
• C. \(10\) cm
• D. \(15\) cm

Hint:

A thin plate over one slit shifts the fringe pattern by a distance of (mu - 1) t D / d.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A thin transparent plate over one slit adds an extra optical path \((\mu - 1)t\), which shifts the whole fringe pattern on the screen.

Step 2: Formula
Shift \(= \dfrac{D}{d}(\mu - 1)t\)
Here the slit separation is \(d = 1\) mm and \(D = 2\) m.

Step 3: Calculation
\(\text{Shift} = \dfrac{2}{1\times 10^{-3}} \times (1.5 - 1) \times 0.06\times 10^{-3}\)
\(= 2000 \times 0.5 \times 6\times 10^{-5} = 0.06\) m

Step 4: Compare the options
\(0.06\) m \(= 6\) cm, which is option (B).

Final Answer:
The fringes are displaced laterally by 6 cm. \[ \boxed{\text{(B) }\text{6 cm}} \]