1.

Air has refractive index 1.0003. The thickness of air column, which will have one more wavelength of yellow light (6000 å) than in the same thickness of vacuum is:- 2 cm 2 mm 2 m 2 km

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FREQUENCY of yellow light in air:

f1=vλ1Here,visvelocityofyellowlight in airλ1iswavelengthoftheyellowlight in air

Frequency of yellow light in vacuum:

f2=cλ2Here,cisvelocityofyellowlight in vacuumλ2iswavelengthoftheyellowlight in vacuum

Equate FREQUENCIES in air and vacuum:

vλ1=cλ2cv=λ2λ1But, cv=μaHere, μa is refractive index of airμa=λ2λ1So, λ1=λ2μa

Number of wavelengths in each medium:

In air, n1=tλ1
In vacuum, n2=tλ2

Here, t is thickness of each column

Difference of thickness of each column:

n1−n2=1tλ1−tλ2=1t(1λ1−1λ2)=1tλ2(λ2λ1−1)=1tλ2(μa−1)=1t=λ2μa−1 =6000×10−8 cm1.0003−1 =0.2 cm  =2 mm
 



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