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Wave Optics

Question
CBSEENPH12039015

Answer the following questions:

(a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.

(b) Light of wavelength 5000 Å propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected? 

Solution

Wavelength of light, straight lambda equals straight space 600 straight space nm straight space equals straight space 600 straight space cross times straight space 10 to the power of – 9 end exponent straight space straight m 

Angular width of the fringe, straight theta equals space 0.1 degree space

Using the formula, theta space equals fraction numerator space lambda over denominator d end fraction
Spacing between the slits, straight d straight space equals fraction numerator straight space straight lambda over denominator straight theta straight space end fraction space equals fraction numerator 600 cross times 10 to the power of negative 9 end exponent over denominator bold italic pi over 180 cross times 0.1 end fraction
                                     equals space fraction numerator 600 cross times 10 to the power of negative 9 end exponent cross times 180 cross times 10 over denominator 3.14 end fraction
                                     equals straight space 34394.90 straight space cross times straight space 10 to the power of – 8 end exponent

equals straight space 0.343 straight space cross times straight space 10 to the power of – 3 end exponent straight space straight m

equals 0.343 straight space cross times straight space 10 to the power of – 3 end exponent straight m
b) Wavelength of light, straight lambda equals straight space 5000 straight space straight A with straight o on top  

Frequency of reflected and refracted light is going to be the same.

Therefore,

Frequency, straight nu straight space equals straight space straight c over straight lambda 
                 equals space fraction numerator 3 cross times 10 to the power of 8 over denominator 5000 cross times 10 to the power of negative 10 end exponent end fraction

italic equals fraction numerator italic 3 italic cross times italic 10 to the power of italic 8 over denominator italic 5 italic cross times italic 10 to the power of italic minus italic 7 end exponent end fraction

equals space italic 3 over italic 5 cross times 10 to the power of italic 15 italic space end exponent

equals space italic 30 over italic 5 cross times 10 to the power of italic 14
equals space 6 cross times 10 to the power of 14 space bold italic H bold italic z
This is the required frequency of both refracted and reflected light. 

Refractive index of water, straight mu space equals space fraction numerator Speed straight space of straight space light straight space in straight space air over denominator Speed straight space of straight space light straight space in straight space water end fraction
4 over 3 equals straight space fraction numerator 3 cross times 10 to the power of 8 over denominator straight v end fraction

space space straight v space equals space fraction numerator 3 cross times 10 to the power of 8 cross times 3 over denominator 4 end fraction space equals space 2.25 space cross times 10 to the power of 8 straight m divided by straight s
Therefore, wavelength of the refracted light,
                  straight lambda to the power of apostrophe space equals space fraction numerator 2.25 space cross times 10 to the power of 8 over denominator 6 cross times 10 to the power of 14 end fraction space equals space 0.375 cross times 10 to the power of negative 6 end exponent straight m