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3 years ago

TRUE OR FALSE: In single-slit diffraction, the central band gets thinner as the width of the slit increases.

Physics

2 answers:

netineya [11]3 years ago

7 0

Answer:

TRUE

Explanation:

As we know that position of minimum intensity on the screen at a given angle of the screen is given by equation

$a sin\theta = N\lambda$

now for width of central maximum we will find the angular width of first minimum

$\beta = \frac{2\lambda}{a}$

now linear width of the central bright fringe is given as

$w = L \beta$

$w = \frac{2\lambda L}{a}$

so we can say that if width of the central fringe will decrease by increasing the width of the slit as they as inversely dependent on each other

So its TRUE

olganol [36]3 years ago

4 0

Answer:

Explanation:

True

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nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

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3 years ago

Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of th

Marysya12 [62]

Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

distance traveled is double which is equal to 2 x 10⁻⁴ cm

a) Number of wave

$N = \dfrac{d}{\mu\lambda}$

$N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}$

N = 2.91

N = 3

b) phase difference is equal to

Reflection from the first surface has a 180° (½λ) phase change.

There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.

net phase difference = $180^0\times \dfrac{3}{2}$

= 270°

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3 years ago

What must crystalline solids have

liberstina [14]

Crystalline solids must have a specific, orderly arrangement of atoms to be considered so.

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3 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

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3 years ago

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

8090 [49]

Answer:

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass c.m= 1.6m

hence 1kg is 1.6m from the c.m

and x is 0.4m from the c.m

Taking moment about the c.m

clockwise moment equals anticlockwise moments

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0

3 years ago