Question

In YDSE distance between the slits and screen is $1.5\text{m}$. When light of wavelength $500\text{nm}$ is used, the $2^{nd}$ bright fringe is obtained on the screen at a distance of $10\text{mm}$ from the central bright fringe. What will be shift in position of $2^{nd}$ bright fringe, if light of wavelength $550\text{nm}$ is used ?

• A. $2\text{mm}$
• B. $1\text{mm}$
• C. $1.5\text{mm}$
• D. $1.1\text{mm}$

Answer 1

The correct option is B $1\text{mm}$

Given
$D=1.5\text{mm}{\lambda }_1=500\text{nm},y_1=10\text{mm},$
${\lambda }_2=550\text{nm},y_2=?$
As $y\propto \lambda$
$\Rightarrow \frac{y_1}{y_2}=\frac{{\lambda }_1}{{\lambda }_2}\Rightarrow \frac{10}{y_2}=\frac{500}{550}$
$\Rightarrow y_2=11\text{mm}\Rightarrow y_2-y_1=1\text{mm}$