Home
Class 12
PHYSICS
A prism of refractive index n(1) & anoth...

A prism of refractive index `n_(1)` & another prism of reactive index `n_(2)` are stuck together without a gap as shown in the figure.The angle of the prisms are as shown. `n_(1)&n_(2)`depend on`lambda`,the wavelength of light according to `n_(1)=1.20+(10.8xx10^(4))/(lambda^(2)) & n_(2)=1.45+(1.80xx10^(4))/(lambda^(2))` where `lambda` is in nm.

(i)Calculate the wavelength `lambda_(0)`for which rays incident at any angle on the interface`BC`pass through without bending at that interface.
(ii) for light of wavelength `lambda_(0)`,find the angle of incidencei on face`AC`such that the deviation produced by the combination of prism is minimum.

Text Solution

Verified by Experts

The correct Answer is:
(a) 600 nm (b) `sin^(-1) (3//4)`

`n_(1) = 1.20 + (10.8 xx 10^(4))/(lambda^(2))` and
`n_(2) = 1.45 + (1.80 xx 10^(4))/(lambda^(2))` Here , `lambda` is in nm.
(a) The incident ray will not deviate at BC if `n_(1) = n_(2)`
`implies 1.20 + (10.8 xx 10^(4))/(lambda_(0)^(2))` = 1.45 + (1.80 xx 10^(4))/(lambda_(0)^(2)) (lambda = lambda_(0))`
`implies (9 xx 10^(4))/(lambda_(0)^(2)) = 0.25 implies lambda_(0) = (3 xx 10^(2))/(0.5) "or" lambda_(0) = 600nm`
(b) The given system is a part of an equiolateral prism of prism angle `60^(@)` as shown in the figure .
At minimum deviation
`r_(1) = r_(2) = (60^(@))/(2) = 30^(@)` = r (say)
`therefore n_(1) = ("sini")/("sinr")`
`therefore "sini" = n_(1)"sin"30^(@)`
`"sin i" = {1.20 + (10.8 xx 10^(4))/((600)^(2))}((1)/(2)) = (1.5)/(2) = (3)/(4)`
`(lambda = lambda_(0) = 600 "nm")`
`implies i = sin^(-1) (3//4)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

A prism of refractive index 1.53 is placed in water of refractive index 1.33 . If the angle of prism is 60^@ , calculate the angle of minimum deviation in water.

A prism of refractive index sqrt(2) has refractive angle 60^(@) . Find the angle of incidence so that a ray minimum deviation.

Knowledge Check

  • For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of prism. Then the angle of the prism is

    A
    `60^@`
    B
    `70^@`
    C
    `50^@`
    D
    none of these
  • For right-angled prism, ray-1 is the incident ri and ray-2 is the emergent ray as shown in the figure. Refractive index of the prism is .........

    A
    `(1)/(sqrt2)`
    B
    `(sqrt3)/(2)`
    C
    `(2)/(sqrt3)`
    D
    `sqrt2`
  • A prism of refractive index mu_1 = 1.42 and prism angle 10° is arranged with another prism of refractive index mu_2 = 1.7. If this combination gives the dispersion without deviation, then what should be the prism angle of second prism?

    A
    `6^@`
    B
    `8^@`
    C
    `10^@`
    D
    `4^@`
  • Similar Questions

    Explore conceptually related problems

    For small prism angle, obtain D_m=(n_(21)-1)

    A ray of light passes through a prism in a principle plane the deviation being equal to angle of incidence which is equal to 2alpha . It is given that alpha is the angle fo prism and mu is the refractive index of the material of prism, then

    A prism of refractive index mu = 1.5 and prisim angle 15^@ is arranged with another prism ( refractive index mu_2 = 1.75. If this combination gives the dispersion without deviation, the what should be the prism angle of second prism ?

    Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then the angle of prism is ...... (sin 48^@36. = 0.75)

    A light of ray incidents normally on one side of equilateral prism. If refractive index of prism is 1.5, then deviation angle is ......