A  beam of light consisting of 2 wavelengths , 650 nm  and 520 nm is used to obtain interference in YDSE . If bright fringes due to both the wavelengths coincide at any point P then least distance of P from central maxima is (slit separation is 2 mm and distance b/w slits and screen is 1.2 m)
1) 1.56 mm
2) 2.36 mm
3) 3 mm
4) 1.8 mm
Please explain the answer in detail.

Asked by Kb Aulakh | 25th Apr, 2015, 06:10: PM

Expert Answer:

Let the wavelength of the two beams be:
begin mathsize 12px style straight lambda subscript space 1 space end subscript equals 650 space nm space equals space 650 cross times 10 to the power of negative 9 end exponent space straight m straight lambda subscript 2 space equals space 520 space nm space equals space 520 space cross times 10 to the power of negative 9 end exponent straight m Given space slit space seperation comma space straight d space equals space 2 space mm space equals space 2 space cross times 10 to the power of negative 3 end exponent space straight m space Distance space between space slit space and space screen comma space straight D space equals space 1.2 space straight m For space the space least space distance space of space coincidence space of space fringes comma space there space must space be space straight a space difference space of space 1 space in space order space of space straight lambda subscript 1 space and space straight lambda subscript 2 straight n space straight beta subscript 1 equals left parenthesis space straight n plus 1 right parenthesis space straight beta subscript 2 fraction numerator straight n space straight D space straight lambda subscript 1 over denominator straight d end fraction space equals space fraction numerator left parenthesis straight n plus 1 right parenthesis space straight D space straight lambda subscript 2 over denominator straight d end fraction straight n space straight lambda subscript 1 space equals left parenthesis straight n plus 1 right parenthesis space straight lambda subscript 2 space straight n space equals fraction numerator straight lambda subscript 2 over denominator straight lambda subscript 1 space minus straight lambda subscript 2 space end fraction straight n equals fraction numerator 520 space cross times 10 to the power of negative 9 end exponent straight m over denominator left parenthesis 650 space minus 520 right parenthesis cross times 10 to the power of negative 9 end exponent space straight m end fraction equals 4 straight n space equals space 4 space Therefore space least space distance comma space fraction numerator straight n space straight D space straight lambda subscript 1 over denominator straight d end fraction equals fraction numerator 4 cross times 1.2 cross times 650 cross times 10 to the power of negative 9 end exponent over denominator space 2 space cross times 10 to the power of negative 3 end exponent end fraction space equals 1.56 cross times 10 to the power of negative 3 end exponent space straight m equals 1.56 space mm Correct space option space colon space left parenthesis 1 right parenthesis space minus space 1.56 space mm end style
The least distance from central maxima is 1.56 mm

Answered by Jyothi Nair | 27th Apr, 2015, 03:17: PM