Question

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?

Answer 1

Answer:

$\lambda =533.6 nm$

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

$dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda$

So, first minima, n = 0

$dsin(\theta_1) = \frac{1}{2}\lambda$

fifth minima, n = 4

$dsin(\theta_5) = \frac{9}{2}\lambda$

$d(sin(\theta_5) -sin(\theta_1))= 4\lambda$

For small angle

$d(tan(\theta_5) -tan(\theta_1))= 4\lambda$

From the figure:

$d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda$

$\frac{d}{D} (y_5-y_1) = 4\lambda$

$\lambda = \frac{d}{4D} (y_5-y_1)$

$\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})$

$\lambda =533.6 nm$

Answer 2

"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:

→ $d (sin \Theta) = (n+\frac{1}{2} ) \lambda$

So,

1st minima, n = 0

→ $d sin(\Theta_1)= \frac{1}{2} \lambda$

5th minima, n = 4

→ $d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda$

For small angle,

→ $d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda$

From the figure, we get

→ $d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda$

  $\frac{d}{D} (y_5 -y_1) = 4 \lambda$

                 $\lambda = \frac{d}{4D}(y_5-y_1)$

                    $= \frac{0.021}{4\times 61} (6.2\times 10^{-3})$

                    $= 533.6 \ nm$

Thus the above answer is correct.  

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