Question

A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe

Answer 1

Answer:

$y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}$

Explanation:

Given that

Wavelength $\lambda=588 \mathrm{nm}$

slit separation $\mathrm{d}=2464 \mathrm{nm}$

slit screen distance $\mathrm{D}=11 \mathrm{cm}$

We know that for double slit the maxima condition is that

$\operatorname{dsin} \theta=m \lambda$

$\sin \theta=\frac{m \lambda}{d}$

$\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)$

For small angle approximation, $\sin \theta \approx \tan \theta \approx \theta$

$\tan \theta=\frac{y_{m}}{D}$

$y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right]$

Now $y_{4}$ $y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}$

Again $y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}$

Hence $y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}$