Question

you pass 633-nm laser light through a narrow slit and observe the diffraction pattern on a screen 6 m away. the distance on the screen between the centers of the first minima on either side of the central bright fringe is 32 mm. how wide is the slit? the intensity at the center of the pattern is i0. what is the intensity at a position on the screen 3 mm from the center of the pattern?

Answer 1

The slit width is approximately
`\(0.002 \ \text{m}\).`The intensity
`\(I(x)\) at \(3 \ \text{mm}\)` from the center is similar to
`\(I_0\)` due to small angle approximation in single-slit diffraction.

The diffraction pattern produced by a single slit can be described by the single-slit diffraction formula, which gives the angular position of the minima:

`\[ \sin \theta_m = m \lambda / b \]`

where:

-
`\( \theta_m \)` is the angle of the
`\( m \)-th`minimum,

-
`\( m \)` is the order of the minimum (for the first minimum,
`\( m = 1 \))`,

-
`\( \lambda \)` is the wavelength of the light,

-
`\( b \)`is the slit width.

The distance between the centers of adjacent minima on the screen is related to the angle
`\( \theta_m \)`and the distance from the slit to the screen (\( L \)) by the equation:

`\[ d_m = \frac{{2L \lambda}}{{b}} \sin \theta_m \]`

In this case, the given values are:

- Wavelength
`\( \lambda = 633 \ nm = 6.33 * 10^(-7) \ m \)`

- Distance to the screen
`\( L = 6 \ m \)`

- Distance between minima
`\( d_m = 32 \ mm = 0.032 \ m \)`

For the first minimum
`(\( m = 1 \)),` we can use the above equations to find the slit width
`(\( b \)):`

`\[ \sin \theta_1 = \frac{{\lambda}}{{b}} \]`

`\[ b = \frac{{\lambda}}{{\sin \theta_1}} \]`

`\[ b = \frac{{\lambda}}{{\sin \left(\arcsin\left(\frac{{d_1}}{{2L\lambda}}\right)\right)}} \]`

`\[ b = \frac{{\lambda}}{{\sqrt{{\frac{{d_1}}{{2L\lambda}}^2}}}} \]`

`\[ b = \frac{{\lambda}}{{\sqrt{{\frac{{d_1^2}}{{4L^2\lambda^2}}}}}} \]`

`\[ b = \frac{{\lambda}}{{\frac{{d_1}}{{2L\lambda}}}} \]`

`\[ b = \frac{{2L\lambda^2}}{{d_1}} \]`

Now, plug in the values:

`\[ b = \frac{{2 * 6 \ \text{m} * (6.33 * 10^(-7) \ \text{m})^2}}{{0.032 \ \text{m}}} \]`

`\[ b \approx 0.002 \ \text{m} \]`

Now, to find the intensity at a position on the screen
`\(3 \ \text{mm}\)`from the center of the pattern
`(\( x = 0.003 \ \text{m} \)),` you can use the intensity formula for single-slit diffraction:

`\[ I(x) = I_0 \left(\frac{{\sin(\beta)}}{{\beta}}\right)^2 \]`

where
`\( I_0 \)` is the intensity at the center of the pattern, and
`\( \beta = \frac{{\pi b x}}{{\lambda L}} \).`

`\[ \beta = \frac{{\pi * 0.002 \ \text{m} * 0.003 \ \text{m}}}{{6.33 * 10^(-7) \ \text{m} * 6 \ \text{m}}} \]`

`\[ \beta \approx 0.03 \]`

Now, calculate
`\( I(x) \):`

`\[ I(x) = I_0 \left(\frac{{\sin(0.03)}}{{0.03}}\right)^2 \]`

Since
`\( \sin(0.03) \approx 0.03 \)`, the intensity will be close to
`\( I_0 \).`Therefore, the intensity at a position
`\(3 \ \text{mm}\)` from the center of the pattern will be approximately the same as the intensity at the center,
`\( I_0 \).`