Question

For "PHYSICS NEWTON"

✯ QUESTION:-

In a double slit experiment with sodium light of wavelength 5893 Å, produces interference pattern on a screen 100 cm away. If the distance between the slits is 0.49 mm, find the distance of
(i) the 10th Bright fringe from the centre;
(ii) 5th Dark fringe from the centre​

Answer 1

Answer:-

$\pink{\bigstar}$ The distance of 10th bright fringe from the centre is $\large\leadsto\boxed{\tt\purple{12 \: mm}}$

$\pink{\bigstar}$ The distance of 5th dark fringe from the centre is $\large\leadsto\boxed{\tt\purple{5.4 \: mm}}$

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• Given:-

• Wavelength = 5893 Å

• Interference pattern on screen 100 cm away.

• Distance between the slits = 0.49 mm

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• To Find:-

• Distance of 10th bright fringe from the centre.

• Distance of 5th dark fringe from the centre.

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• To Know:-

• 1 Å = 1 × 10^-10 m

• 1 mm = 10-³ m

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• Solution:-

We know,

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✯ The distance of nth bright fringe from the central is

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{x = \dfrac{n \lambda D}{d}}}}$

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• Substituting in the Formula:-

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➪ $\sf x = \dfrac{10 \times 5893 \times 10^{-10} \times 1}{0.49 \times 10^{-3}}$

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➪ $\sf x = \dfrac{5893 \times 10^{-9}}{0.49 \times 10^{-3}}$

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➪ $\sf x = \dfrac{5893 \times 10^{-6}}{0.49}$

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➪ $\sf x = 12026.5 \times 10^{-6}$

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➪ $\sf x = 12.0\times 10^{-3}$

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★ $\large{\bf\red{x = 12 \: mm}}$

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Therefore, the distance of 10th bright fringe from the centre is 12 mm.

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Now,

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✯ The distance of nth dark fringe from the central is

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$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{x = (2n - 1) \dfrac{n \lambda D}{2d}}}}$

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• Substituting in the Formula:-

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➪ $\sf x = \dfrac{(2 \times 5 - 1) \times 5893 \times 10^{-10} \times 1}{2 \times 0.49 \times 10^{-3}}$

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➪ $\sf x = \dfrac{9 \times 5893 \times 10^{-10}}{0.98 \times 10^{-3}}$

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➪ $\sf x = \dfrac{53037 \times 10^{-7}}{0.98}$

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➪ $\sf x = 54119.38 \times 10^{-7}$

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➪ $\sf x = 5.4 \times 10^{-3}$

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★ $\large{\bf\red{x = 5.4 \: mm}}$

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Therefore, the distance of 5th dark fringe from the centre is 5.4 mm.

Answer 2

To Find:-

• Find the 10th Bright fringe from the centre.

• Find the 5th fringe from the centre.

Given:-

• Sodium light of wavelength 5893 Å .
• Produces interference pattern on a screen 100 cm away.
• Distance between the slits is 0.49 mm .

Solution:-

Here ,

• 1Å = $\tt \: 1 \times { 10 \: m }^{ - 10 }$

• 1 mm = $\tt \: { 10 \: m }^{ - 3 }$

Formula:-

$\large\underline{\boxed{x = \dfrac{n \lambda D}{d}}}$

$\tt\implies \: x = \dfrac { 10 \times 5893 \times { 10 }^{ -10 } \times 1 } { 0.49 \times { 10 } { -3 } }$

$\tt\implies \: x = \dfrac { 5893 \times { 10 }^{ -9 } } { 0.49 \times { 10 } { -3 } }$

$\tt\implies \: x = 12 \times { 10}^{ -3 }$

$\tt\implies \: x = 12 mm$

Hence ,

• Distance is 12 mm .

Now ,

We have to find 5th fringe from the centre:-

Formula:-

$\large\underline{\boxed{x = (2n - 1) \dfrac{n \lambda D}{2d}}}$

$\tt\implies \: x = \dfrac { ( 2 \times 5 - 1 ) \times 5893 \times { 10 }^{ -10 } \times 1 } { 2 \times 0.49 \times { 10 } { -3 } }$

$\tt\implies \: x = \dfrac { 53037 \times { 10 }^{ -7 } } { 0.98 }$

$\tt\implies \: x = 54119.38 \times { 10 }^{ - 7 }$

$\tt\implies \: x = 5.4 \: m \: m$

Hence ,

• Distance is 5.4 mm