Home
Class 11
PHYSICS
Refractive index mu is given as mu=A+B/l...

Refractive index mu is given as `mu=A+B/lambda^2,` where A and B are constants and lambda is wavelength, then dimensions of B are same as that of

A

Wavelength

B

Area

C

Volume

D

Velocity

Text Solution

Verified by Experts

The correct Answer is:
b

The refractive index has no dimensions.
`therefore (B)/(lambda^(2))` has no dimensions.
i.e. B has the same dimensions as that of `lambda^(2)` or `("Length")^(2)` or (Area).
Promotional Banner

Topper's Solved these Questions

  • MEASUREMENTS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos
  • MAGNETISM

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos
  • RAY OPTICS (MIRRORS, LENSES AND OPTICAL INSTRUMENTS)

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos

Similar Questions

Explore conceptually related problems

The refractive index of a medium is given by mu=A+(B)/(lambda^(2)) where A and B are constants and lambda is the wavelength of light. Then the dimensions of B are the same as that of

Find the dimension of B^2/(2 mu_0)

(i) mu=(c)/(v_m)=(vlambda)/(vlambda_m) where c: speed of light in vacuum v_m: speed of light in medium lambda: wavelength in vacuum lambda_m : wavelength in medium upsilon: frequency of light (ii) mu=((epsilonmu)/(epsilon_0mu_0))^(1/2) where epsilon: electric permittivity in free space epsilon: electric permittivity in medium mu_0: magnetic permeability in free space mu: magnetic permeability in medium (iii) mu=A+B/lambda^2 , where A and B are constants (iv) In case of VIBGYOR, wavelength increases from violet to red, refractive index decreases from violet to red. Wavelength for red-colour is maximum whereas refractive index is maximum for violet colour.

The momentum of an electron in an orbit is h//lambda , where h is a constant and lambda is wavelength associated with it. The nuclear magneton of electron of charge e and mass m_(e) is given as mu_(n)=(eh)/(3672 pim_(e)) . The dimension of mu_(n) are (Ararr"current")

The angles of incidence and refraction of a monochromatic ray of light of wavelength lamda at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread delta lamda in wavelength about a mean wavelength lamda is refracted at the same air-glass interface. The refractive index mu of glass depends on the wavelength lamda as mu(lamda) = a + b//lamda2 where a and b are constants. Then the angular spread in the angle of refraction of the beam is

The refractive index mu of a medium is found to vary with wavelength lambda as mu = A +(B)/(lambda^2). What are the dimensions of A and B?

Determine what happens to the double slits interference pattern if one of the slits is covered with a thin, transparent film whose thickness is lambda/(2(mu-1)) , where lambda is the wavelength of the incident light and mu is the index of refraction of the film.

A prism of a refracting angle 60^(@) is made with a material of refractive index mu . For a certain wavelength of light, the angle of minimum deviation is 30^(@) . For this wavelength, the value of mu of material is

When light of wavelength lambda travels through glass, the refractive index of glass varies with wavelength as mu=A+(B)/(lambda^(2)) . What are the dimensional formulae of A and B?