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sir please answer this question

In a young`s double slit experiment'slits are seperated by 0.5mm, and the screen is place 150 cm away. A beam of light consisting of two wavelengths, 650nm and 520 nm, is used to obtain fringes on the screen . the least distance from common central maximum to the point were the bright fringes due to both the wavelngths coincide is

 

Asked by ppavani198 29th December 2018, 6:37 PM
Answered by Expert
Answer:
Let λ1 = 650 nm and λ2 = 520 nm are the two wavelenths present in the light beam.
 
Let bright fringe number n1 of λ1 coincide with fringe number n2 of λ2 .
Let D is the distance between slit and screen and d is interdistance between slits.

Then we have
 
begin mathsize 12px style fraction numerator n subscript 1 lambda subscript 1 D over denominator d end fraction space equals space fraction numerator begin display style n subscript 2 lambda subscript 2 D end style over denominator begin display style d end style end fraction space space space o r space space space n subscript 1 lambda subscript 1 space equals space n subscript 2 lambda subscript 2
h e n c e space n subscript 1 cross times 650 space equals space n subscript 2 cross times 520 space space o r space space space n subscript 1 over n subscript 2 space equals space 520 over 650 space equals 4 over 5
T h e n space w e space c o n s i d e r comma space space space n subscript 1 space equals space 4 space space a n d space space space n subscript 2 space equals space 5 end style
 
Hence the required minimum distance from central maximum is begin mathsize 12px style fraction numerator n subscript 1 lambda subscript 1 D over denominator d end fraction space equals space fraction numerator 4 cross times 650 cross times 10 to the power of negative 9 end exponent cross times 150 cross times 10 to the power of negative 2 end exponent over denominator 0.5 cross times 10 to the power of negative 3 end exponent end fraction space equals space 7.8 space m m end style
Answered by Expert 30th December 2018, 10:42 AM
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