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BartSMP [9]
3 years ago
11

A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. From the center of one p

articular fringe to the center of the ninth bright fringe from this one is 1.6 cm. What is the wavelength of the light? [530nm]
Physics
2 answers:
Alexandra [31]3 years ago
6 0

Answer:

530 nm

Explanation:

Screen distance, D = 1 m

slit distance, d = 0.3 mm

n = 9 th bright

y = 1.6 cm

Let λ be the wavelength of light used.

y = n x D x λ / d

1.6 x 10^-2 = 9 x 1 x λ / (0.3 x 10^-3)

λ = 5.3333 x 10^-7 m

λ = 533.33 nm

λ = 530 nm ( by rounding off)

Pie3 years ago
5 0

Answer:

The wavelength of the light is 530 nm.

Explanation:

Given that,

Distance D= 1.0 m

Distance between slits d= 0.30 mm

Number of fringe = 9

Width = 1.6 cm

We need to calculate the angle

Using formula of angle

\tan\theta=\dfrac{y}{D}

tan\theta=\dfrac{1.6\times10^{-2}}{1.0}

\theta=\tan^{-1}(\dfrac{1.6\times10^{-2}}{1.0})

\theta=0.91^{\circ}

We need to calculate the wavelength of the light

Using formula of wavelength

d\sin\theta=m\lambda

\lambda=\dfrac{d\sin\theta}{m}

Put the value into the formula

\lambda= \dfrac{0.30\times10^{-3}\times\sin0.91}{9}

\lambda=5.29\times10^{-7}\ m

\lambda=530\ nm

Hence, The wavelength of the light is 530 nm.

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