00:01
Hello everyone, we are going to understand this question.
00:04
Here it is given that a camera lens and is equal to 1 .25 is coated with, so here given refractive index of lens, refractive index of lens that is represented by mu1 which is 1 .5 and it is coated with a thin film of magnesium fluoride, refractive index of thin films.
00:48
Refractive index of thin films that is represented by mu2 sorry it is represented by n1 and n2 n1 and n2 and n2 and 2 is 1 .38 thickness is thickness t is 90 .0 nanometer or we can write 90 .0 into 10 to the 4 minus 9 meter so what is the wavelength in the visible spectrum is most strongly transmitted through the film so wavelength that we have to find now now we can write 2 into t is equal to wavelength of film upon 2 from this we will get lambda wavelength in the film, wavelength in the coated film, that is 4 into t, so 4 into 90 nanometer.
02:37
After calculation we will get 360 nanometer.
02:41
Now, now, wave length of light, lambda is equal to lambda film upon n1.
03:03
Here it is lambda film upon reflective index of film, that is, n2.
03:09
So we can write 360 into 10 to the power minus 9 upon value of n2 is 1 .38.
03:21
So after calculation we will get lambda is equal to, after calculation we will get 200 into 10 to 4 minus 9 meter.
03:35
Or we can write lambda is equal to 200 nanometer.
03:41
Lambda is equal to 200 nanometer.
03:46
Is the wavelength of visible life spectrum.
03:51
Here we did a calculation mistake.
03:55
Here we have to find the wavelength of light in air...