A certain day absorbs lights of `lamda=400` nm and then fluorescence light of wavelength `500` nm. Assuming that under given condition `40%` of the absorbed energy is re-emitted as fluorescence, calculate the ratio of quanta absorbed to number of quanta emitted out.

Light of wavelength 500 nm falls on a metal whose work function is 1.9 eV. find the kinetic energy of the photo electrons emitted.

In a Young's double slit experiment,12 fringes are observed to be formed in a certain segment of the screen,when light of wavelength 600 nm is used.If the wavelength of light is changed to 400 nm ,number of fringes observed in the same segment of the screen is given by

[Ti(H_2O)_6]^(3+) absorbs light of a wavelength 500 nm. Name one ligand which would form Ti(IID) complex absorbing light of lower wavelength than 500 nm and one ligand which would form acomplex absorbing light of wavelength higher than 500 nm.

A beam of light of wavelength 400 nm is incident normally on a right angled prism as shown in fir.2.45. . It is observed that the light just grazes long the surface AC after falling on it.Give that there fractive index of the material of the prism varies with the wavelength as per the relation. mu=1.2 + b/(gamma^2) Calculate the value of b and the refractive index of the prism material for a wavelength gamma = 500 nm .Given theta=sin^(-1) 0.625 .

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1nm and the other visible light of 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength?

Infrared lamps are used in restaurants to keep the food warm. The infrared radiation is strongly absorbed by water, raising its temperature and that of the food. If the wavelength of infrared radiationis assumed to be 1500 nm, and the number of quanta of infrared radiation produced per second by an infrared lamp (that consumes enregy at the rate of 100 W and is 12% effcient only) is (xx10^(19) , then the value of x is : (Given: h= 6.665xx10^(-34)J-s)

A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB [Fig.6.07].The intensity of the reflected light is 36% of the incident intensity.Interfrerence fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. If the intensity t pint P corresponds to a maximum,calculate the minimum distance through which the reflecting surface AB sould be shifted so that the intensity at P again becomes maximum.