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:

Solution:-

Here ,

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

•

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:-

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

Answer 2

★ Question Given :

• 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

★ Required Solution :

✰ Formula to be used :

• ➲ x = nλD / d

✰ Putting value in Formula :

[ i ]

• ➢ x = 10 × 5893 × 10 ^-10 × 1 / 0.49 × 10^-3

• ➢ x = 5893 × 10^-9

• ➢ x = 12 × 10^-3

• ➢ x = 12mm

★ Therefore :

• ➥ The distance is = 12mm

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[ ii ]

✰ Finding 5th Dark fringe from the centre :

• ➲ x = (2n-1) nλD / 2d

• ➢ x = (2 × 5 - 1) × 5893 × 10^-10 / 2 × 0.49 × 10-³

• ➢ x = 53037 × 10^7 / 0.98

• ➢ x = 54119.38 × 10^-7

• ➢ x = 5.4mm

★ Therefore :

• ➥ The distance is x = 5.4mm

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