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3 years ago

In Young’s double-slit experiment using monochromatic light, the interference pattern consists of a central _____.

Physics

2 answers:

svetoff [14.1K]3 years ago

8 0

Answer:

B)bright band with alternate dark and bright bands on either side

Explanation:

In young's double slit experiment we verify the wave characteristic of light

In this wave phenomenon we can see that on screen there is path difference of waves coming from two slits of light source due to which the total intensity of light on the screen will change accordingly

At the central point of the screen the path difference of light is ZERO

due to this the net intensity of the light on the central part of the screen will be maximum.

Since it is a monochromatic light experiment so now when we move on either side on the central point the path difference will change and hence we will get the maximum and minimum intensity alternatively on the screen

So correct answer will be

B)bright band with alternate dark and bright bands on either side

Pie3 years ago

5 0

In young's double slit experiment that uses monochromatic light the interference pattern formed has Central bright band with alternate dark and bright band That is option B

Explanation-

In this young experiment two small slits Namely a and b are formed on the screen and a monochromatic light is focused on them. Wavelets come out of of this lets scintillating and overlapping each other. This we get from huyginns principal, thus forming alternate dark and bright bands with bright band at center and all bands are about one meter apart.

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3 years ago

Read 2 more answers

A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits

slavikrds [6]

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

lets first calculate the time it took the fish to fall

s = ut + $(\frac{1}{2}) at^{2}$

since u = 0

s = $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 4.7 }{9.8} }$ = 0.98 s

pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s