Home
Class 12
PHYSICS
The XY plane is the boundary between two...

The XY plane is the boundary between two tranparednt media. Medium 1 with `z ge 0` has a refraxtive index of `sqrt2` and medium 2 with `z le 0` has a refractive index of `sqrt3`. A ray of light in medium 1 given by the vector `6sqrt(3)hat(i)+8sqrt(3)hat(j)-10hatk` is incident on teh plane of separation. Find the unit vector in the direction of teh refracted ray in medium 2.

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(5sqrt2) (3hati + 4 hatj - 5hatk)`
Incident ray
`vecA = 6sqrt3hati + 8sqrt3hatj - 10 hatk = (6sqrt3 hati + 8sqrt3hatj) + (-10)hatk`

`QvecO + PvecQ` (As shown in figure )
Note that `QvecO` is lying on x-y plane .
Note , OQ' and Z-axis are mutually perpendicular .
Hence , we can show them in two- dimensional figure as below .

Vector `vecA` makes an angle i with z-axis , given by
`i = cos^(-1).{(10)/(sqrt((10)^(2) + (6sqrt3)^(2) + (8sqrt3)^(2)))} = cos^(-1){(1)/(2)}`
`i = 60^(@)`
Unit vector in the direction of `QOQ'` will be
`hatq = (6sqrt3hati + 8sqrt3hatj)/(sqrt((6sqrt3)^(2) + (8sqrt3)^(2))) = (1)/(5)(3hati + 4 hatj)`
Snell's law gives `(sqrt3)/(sqrt2) = ("sini")/("sinr") = ("sin"60^(@))/("sinr")`
`therefore "sinr" = (sqrt3//2)/(sqrt3//sqrt2) = (1)/(sqrt2) therefore r = 45^(@)`
Now , we have to find out a unit vector in refracted ray's direction OR . say it is `hatr` whose magnitude is 1 .
Thus `hatr = (1"sinr")hatq - (1 - "cosr") hatk = (1)/(sqrt2)(hatq - hatk)`
`= (1)/(sqrt2) [(1)/(5) (3hati + 4 hatj) - hatk]`
`hatr = (1)/(5sqrt2). (3hati + 4hatj - 5hatk)`
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Let the x-z plane be the boundary between two transparent media. Medium 1 in zge0 has a refractive index of sqrt2 and medium 2 with zlt0 has a refractive index of sqrt3 . A ray of light in medium 1 given by the vector vecA=6sqrt3hati+8sqrt3hatj-10hatk is incident on the plane of separation. The angle of refraction in medium 2 is:

Let the x-z plane be the boundary between two transparent media. Media 1 in zge0 has refractive index of sqrt2 and medium 2 with zlt0 has a refractive index of sqrt3 . A ray of light in medium 1 given by the vector vecA=6sqrt3hati+8sqrt3hatj-10hatk in incident on the plane of separation. The angle of refraction in medium 2 is

The x-y plane is the boundary between two transparent media. Medium-1 with zle0 has a refractive index sqrt(2) and medium -2 with zle0 has a refractive index sqrt(2) . A ray of light in medium -1 given by vector A=sqrt(3)hati-hatk is incident on the plane of separation. The unit vector in the direction of the refracted ray in medium-2 is

[" The "xz" plane is the boundary "],[" between two transparent medium."],[" Medium - I with "y>=0" has "],[" refractive index "mu" and medium - II "],[" with "y<=0" has a refractive index 1."],[" A ray of light in medium-I,given by "],[" vector "vec n=hat i-sqrt(3)hat j" is incident on "],[" the plane of separation.If the "],[" reflected and refracted rays make "],[" an angle of "90^(@)" with each other,"],[" then the value of "mu" is "sqrt(k)" .Find the "],[" value of "k" ."]

A ray of light is incident on a medium of refractive index sqrt(2) at an angle of incidence of 45^(@) . The ratio of the width of the incident beam in air to that of the refracted beam in the medium is

Two plane mirrors are combined to each other as such one is in (y-z) plane and other is in (x-z)plane.A ray of light along vector (1)/(sqrt(3)) hati +(1)/(sqrt(3))hatj +(1)/(sqrt(3))hatk is incident on the first mirror. Find the unit vector in the direction of emergence ray after successive reflections through these mirros.