A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 61 cm behind the slits is 6.2 mm What is the wavelength of the light used in this experiment?

Respuesta :

Answer:

[tex]\lambda =533.6 nm [/tex]

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

[tex]dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda[/tex]

So, first minima, n = 0

[tex]dsin(\theta_1) = \frac{1}{2}\lambda[/tex]

fifth minima, n = 4

[tex]dsin(\theta_5) = \frac{9}{2}\lambda[/tex]

[tex]d(sin(\theta_5) -sin(\theta_1))= 4\lambda[/tex]

For small angle

[tex]d(tan(\theta_5) -tan(\theta_1))= 4\lambda[/tex]

From the figure:

[tex]d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda[/tex]

[tex]\frac{d}{D} (y_5-y_1) = 4\lambda[/tex]

[tex]\lambda = \frac{d}{4D} (y_5-y_1)[/tex]

[tex]\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})[/tex]

[tex]\lambda =533.6 nm[/tex]

"533.5 nm" would be the wavelength of the light used in this experiment.

According to the question,

Slits spacing,

  • d = 0.21 mm

Distance of screen,

  • D = 61 cm

The condition for minima is given as:

→ [tex]d (sin \Theta) = (n+\frac{1}{2} ) \lambda[/tex]

So,

1st minima, n = 0

→ [tex]d sin(\Theta_1)= \frac{1}{2} \lambda[/tex]

5th minima, n = 4

→ [tex]d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda[/tex]

For small angle,

→ [tex]d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda[/tex]

From the figure, we get

→ [tex]d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda[/tex]

  [tex]\frac{d}{D} (y_5 -y_1) = 4 \lambda[/tex]

                 [tex]\lambda = \frac{d}{4D}(y_5-y_1)[/tex]

                    [tex]= \frac{0.021}{4\times 61} (6.2\times 10^{-3})[/tex]

                    [tex]= 533.6 \ nm[/tex]

Thus the above answer is correct.  

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