Question

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

• A. 2mm
• B. 1mm
• C. 1.5mm
• D. 1.1mm

Answer 1

Answer: (B)

1mm

In YDSE,the formula for bright fringes (constructive interference) is:

$y_n=\frac{n\lambda D}{d}$ for nth bright fringe

Initially, $D=1.5m$

$\lambda =500nm=500\times {10}^{-9}m$

$n=2$

$y_2=10mm=10\times {10}^{-3}m={10}^{-2}m$

Substituting: $d=\frac{2\times 500\times {10}^{-9}\times 1.5}{{10}^{-2}}$

$=1500\times {10}^{-7}=1.5\times {10}^{-4}m$

Now,if $\lambda =550nm:$

$y_2^{\prime }=\frac{2\times 550\times {10}^{-9}\times 1.5}{1.5\times {10}^{-4}}$

$=1100\times {10}^{-5}=1.1\times {10}^{-2}m$

Shift $=y_2^{\prime }-y_2=1.1\times {10}^{-2}-1\times {10}^{-2}$

$={10}^{-2}(1.1-1)=0.1\times {10}^{-2}={10}^{-3}=1mm$