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elena55 [62]
4 years ago
9

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

4.6 mm behind the grating. Part A What is the distance between the two mm
Physics
1 answer:
Ksenya-84 [330]4 years ago
8 0

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d=\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m

Wavelength of light,\lambda=510 nm=510\times 10^{-9} m

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

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

For first bright fringe, =1

sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305

\theta=sin^{-1}(0.305)=17.76^{\circ}

The distance between two m=1 fringes

x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

sin\theta=1

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

Substitute the values

1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}

m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3

Maximum number of bright fringes on the scree=2m+1=2(3)+1=7

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Green part of the visible spectrum.

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\lambda_{max}=\frac{2.898\times 10^{-3}}{5800}\\\Rightarrow\lambda_{max}=0.499\times 10^{-6}=0.5\mu m=500\ nm

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So, the wavelength would be of around the X ray part of the electromagnetic spectrum.

Human body temperature = 37°C = 37+273.15 = 310.15 K

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Explanation:

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The time period of a simple pendulum is given by :

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Where

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l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

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Put in equation (1)

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