Question

In a YDSE experiment $\lambda$= 540 nm, D= 1m, d= 1 mm. A thin film is pasted on upper slit and the central maxima shifts to the point just in of front of the upper slit. What is the path difference at the centre of the screen?

• A. 540 nm
• B. 270 nm
• C. 500 nm
• D. 810 nm

Answer 1

The correct option is C 500 nm

$Given:$ $\lambda =540nm$ ; $D=1m$ ; $d=1mm$

$Solution:$ In front of upper slint

On screen $=\Delta x=d\left(\frac{d/2}{D}\right)-(\mu -1)t=0$

$\Delta x=d\frac{\left(d/2\right)}{D}-(\mu -1)t=0$

at center on the screen

$\Delta x=(\mu -1)t=\frac{d^2}{2D}$

$\Delta x=\frac{{10}^{-6}}{2}$

$\Delta x=500\times {10}^{-9}m$

$\Delta x=500nm$

$So,the$ $correct$ $option:C$