Question

In a double-slit interference experiment, light of frequency 6.0 × 10^14 Hz is incident on a pair of slits. Bright fringes that are 3.0 mm apart are observed on a screen some distance away. What is the separation of the bright fringes when the frequency of the light is changed to 5.0 × 10^14 Hz?
A 1.8 mm
B 2.5 mm
C 3.0 mm
D 3.6 mm

Answer 1

Answer:

Explanation:

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Answer 2

The separation of the bright fringes is 3.6 mm .

Option (D) is correct.

Explanation:

Given :

Frequency of light $f_{1} = 6 \times 10^{14}$ Hz

Bright fringe width $\beta_{1} = 3$ mm

Frequency of different light $f _{2} = 5 \times 10^{14}$ Hz

Fringe width is given by,

  $\beta = \frac{ \lambda D}{d}$

Fringe width is directly proportional to the wavelength of light.

Here in example we have only one variable parameter which is frequency.

We know that wavelength is inversely proportional to the frequency,

   $c = f \lambda$ ,  $\lambda$ ∝ $\frac{1}{f}$

So we can write relation between fringe width and frequency,

      $\beta _{1} f_{1} = \beta_{2} f_{2}$

    $\beta _{2} = \frac{3 \times 6 \times 10^{14} }{5 \times 10^{14} }$                     ( in mm )

    $\beta _{2} = 3.6$ mm

Thus, the separation of the bright fringes is 3.6 mm