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Physics: Post your doubts here!

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Can someone explain to me what would happen if white light is incident on a diffraction gratinginstead of a monochromatic light?

PLEASE can u explain it thanx a lot!!!!
 
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Can someone explain to me what would happen if white light is incident on a diffraction gratinginstead of a monochromatic light?

PLEASE can u explain it thanx a lot!!!!

You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
 
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Can any solve this:
The frequency of the fundamental mode of transverse vibration of stretched string wire 1 m long is 250 Hz.when the wire is shortened to 0.4 m at the same tension the fundamental frequency is?

A: 102 Hz
B: 162 Hz
C: 312 Hz
D: 416 Hz
E: 640 Hz
 
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You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
thanx alot again!!!
 
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Can any solve this:
The frequency of the fundamental mode of transverse vibration of stretched string wire 1 m long is 250 Hz.when the wire is shortened to 0.4 m at the same tension the fundamental frequency is?

A: 102 Hz
B: 162 Hz
C: 312 Hz
D: 416 Hz
E: 640 Hz
fundamental wavelength =2L
L = 1/2 lambda
lambda = 2L
V is constant both cases, so f1(lambda1) = f2(lambda2)
(250)(2) = 0.8f
f = 625
answer is 625 i had contacted many phds uni teachers and students they all agree with this answer
 
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You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
do u agree?
with my answer as 625 ??
 
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fundamental wavelength =2L
L = 1/2 lambda
lambda = 2L
V is constant both cases, so f1(lambda1) = f2(lambda2)
(250)(2) = 0.8f
f = 625
answer is 625 i had contacted many phds uni teachers and students they all agree with this answer
I even got the answer as 625 but there was a difference of 15 in the mcQ's answer and my answer so i might think its wrong:
Anyways thnx brood:)
 
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What is the detailed solution of:-
View attachment 22028

let energy before rebounding E1 and energy after rebounding E2
energy is proportional to mgh
m and g are constant here they arent changed

since m and g are constants dividing E2 by E1 we get 0.5mgh/mgh=1/2
kinetic energy=(1/2)mv^2
now lets go kinetic energy part (1/2)m is constant here


kinetic energy=1/2mu^2 before rebounding
kinetic energy=1/2mv^2 after rebounding
ratio of k.e after rebounding/ k.e before rebounding=v^2/u^2
before rebounding potiential energy lose and kinectic energy gain
after rebounding potiential energy gain and kinetic energy lose but remember energy is conserved
p.e is converted into k.e
and k.e into p.e
so we can say kinetic energy before rebounding equals E1
and k.e before rebounding equals E2

since we want ratio of u/v
so
v^2/u^2=1/2
square rooting both sides we get
v/u=1/sqrt(2)
 
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please please help me with b(i) and (ii), i was not able to draw both figures in the graph as well as on the diagram. I could not draw using the guidelines given in the marking scheme., so i would be thankful if anyone drew them for me.
The questions are uploaded.
 

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