Question

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Answer 1

Given:

Plane wavefront of light of wavelength 5500 A° is incident on two slits in a screen perpendicular to the direction of light rays. Separation of 10 bright fringes is 2 cm.

To find:

Distance between slits ?

Calculation:

Total separation of 10 bright fringes = 2 cm, so seperation is 2 adjacent fringes = 2/10 = 0.2 cm.

So, our fringe width = $\beta$ = 0.2 cm.

Now, let distance between slits be d :

• All quantities put in SI units.
• D = 2m

$\therefore \: \beta = \dfrac{ \lambda D}{d}$

$\implies \: d = \dfrac{ \lambda D}{ \beta }$

$\implies \: d = \dfrac{5500 \times {10}^{ - 10} \times 2}{0.2 \times {10}^{ - 2} }$

$\implies \: d = \dfrac{5500 \times {10}^{ - 10} }{0.1 \times {10}^{ - 2} }$

$\implies \: d = \dfrac{5500 \times {10}^{ - 10} }{ {10}^{ - 3} }$

$\implies \: d = 5500 \times {10}^{ -7} \: m$

$\implies \: d = 0.55 \times {10}^{ -3} \: m$

$\implies \: d = 0.55 \: mm$

So, distance between slits is 0.55 mm.

Answer 2

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