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

`y_(4)-y_(3)=35.22-11.27=23.95 \mathrm{cm}`

Step-by-step 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

`\sin \theta=(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=(y_(m))/(D)`

`y_(m)=D * \tan \left[\sin ^(-1)\left((m \lambda)/(d)\right)\right]`

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

Again
`y_(3)=D * \tan \left[\sin ^(-1)\left((3 \lambda)/(d)\right)\right]=11 * \tan \left[\sin ^(-1)\left(\frac{3 * 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}`