Question

If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away

Answer 1

Answer:

0.82 mm

Explanation:

The formula for calculation an $n^{th}$ bright fringe from the central maxima is given as:

$y_n=\frac{n \lambda D}{d}$

so for the distance of the second-order fringe when wavelength $\lambda_1$ = 745-nm can be calculated as:

$y_2 = \frac{n \lambda_1 D}{d}$

where;

n = 2

$\lambda_1$ = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

$y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y_2$ = 0.00276 m

$y_2$ = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength $\lambda_2$ = 660-nm is as follows:

$y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y^'}_2$ = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

$\delta y = y_2-y^{'}_2$

$\delta y$ = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

$\delta y$ = 10 ⁻³ (2.76 - 1.94)

$\delta y$ = 10 ⁻³ (0.82)

$\delta y$ = 0.82 × 10 ⁻³ m

$\delta y$ =  0.82 × 10 ⁻³ m $(\frac{1.0mm}{10^{-3}m} )$

$\delta y$ = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm