Please solve this question.
Please solve this question. Q3.
A monochromatic light of wavelength A1 travelling in a medium of refractive index
enters a denser medium of refractive index Y2. The wavelength in the second medium
is:
A.
A1 (P2 ¯ PI)/ß2
B.
D.
Dear friend,
When monochromatic light moves from one medium to another, it's wavelength changes whereas frequency remains same.
--> frequency = same
--> speed (1st medium) = V1
--> speed (2nd medium) = V2
Wavelength in 1'st medium = lambda1
We have to find wavelength in second medium, i.e, lambda2
Refractive index of 1'st medium = U1
------> U1 = C/V1 ----(1)
Refractive index of 2'nd medium = U2
------> U2 = C/V2 ----(2)
Relation between frequency, wavelength and speed of light is :-
---> Speed = frequency * Wavelenth
In rarer medium
---> V1= f1 * lambda1
---> f1 = V1/lmbda1
Similarly :-
---> V2 = f2 * lambda2
---> f2 = V2/lambda2
Since frequency remains same, so
---> f1 = f2
Therefore :-
V1/lambda1 = V2/lambda2 ----(3)
Now,
---> V1 = C/U1
---> V2 = C/U2
Putting the values in equation (3)
--> (C/U1)/lambda1 = (C/U2)/lambda2
--> C/U1 * 1/lambda1 = C/U2 * 1/lambda2
C will be cancelled on both the sides
--> lambda2 = (U1*l ambda1)/U2
--> lambda2 = lambda1(U1/U2)
Hence option A is correct !
Hope it helped !
And Don't forget to hit thumbs up if it really helped !
Thanks and Regards !
Abhay Pratap Singh......(^^)
When monochromatic light moves from one medium to another, it's wavelength changes whereas frequency remains same.
--> frequency = same
--> speed (1st medium) = V1
--> speed (2nd medium) = V2
Wavelength in 1'st medium = lambda1
We have to find wavelength in second medium, i.e, lambda2
Refractive index of 1'st medium = U1
------> U1 = C/V1 ----(1)
Refractive index of 2'nd medium = U2
------> U2 = C/V2 ----(2)
Relation between frequency, wavelength and speed of light is :-
---> Speed = frequency * Wavelenth
In rarer medium
---> V1= f1 * lambda1
---> f1 = V1/lmbda1
Similarly :-
---> V2 = f2 * lambda2
---> f2 = V2/lambda2
Since frequency remains same, so
---> f1 = f2
Therefore :-
V1/lambda1 = V2/lambda2 ----(3)
Now,
---> V1 = C/U1
---> V2 = C/U2
Putting the values in equation (3)
--> (C/U1)/lambda1 = (C/U2)/lambda2
--> C/U1 * 1/lambda1 = C/U2 * 1/lambda2
C will be cancelled on both the sides
--> lambda2 = (U1*l ambda1)/U2
--> lambda2 = lambda1(U1/U2)
Hence option A is correct !
Hope it helped !
And Don't forget to hit thumbs up if it really helped !
Thanks and Regards !
Abhay Pratap Singh......(^^)
- 7
When monochromatic light moves from one medium to another, it's wavelength changes whereas frequency remains same.
--> frequency = same
--> speed (1st medium) = V1
--> speed (2nd medium) = V2
Wavelength in 1'st medium = lambda1
We have to find wavelength in second medium, i.e, lambda2
Refractive index of 1'st medium = U1
------> U1 = C/V1 ----(1)
Refractive index of 2'nd medium = U2
------> U2 = C/V2 ----(2)
Relation between frequency, wavelength and speed of light is :-
---> Speed = frequency * Wavelenth
In rarer medium
---> V1= f1 * lambda1
---> f1 = V1/lmbda1
Similarly :-
---> V2 = f2 * lambda2
---> f2 = V2/lambda2
Since frequency remains same, so
---> f1 = f2
Therefore :-
V1/lambda1 = V2/lambda2 ----(3)
Now,
---> V1 = C/U1
---> V2 = C/U2
Putting the values in equation (3)
--> (C/U1)/lambda1 = (C/U2)/lambda2
--> C/U1 * 1/lambda1 = C/U2 * 1/lambda2
C will be cancelled on both the sides
--> lambda2 = (U1*l ambda1)/U2
--> lambda2 = lambda1(U1/U2)
Hence option A is correct !
- 1