Question

Two wavelengths of sodium light 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2×10
−6
m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

Answer 1

Answer:

Explanation:

For first maximum,

y  

1

​  

=  

a

λD

​  

 

For 590 nm,

y  

1,590

​  

=  

2×10  

−6

 

590×10  

−9

×1.5

​  

 

y  

1,590

​  

=0.4425m

For 596 nm,

y  

1,596

​  

=  

2×10  

−6

 

596×10  

−9

×1.5

​  

 

y  

1,596

​  

=0.447m

Separation is

y  

1,596

​  

−y  

1,590

​  

=0.0045m

Answer 2

Answer:

$6.75 \: mm$

Explanation:

Wavelength of first Sodium line, λ1 = 590 nm

Wavelength of second sodium line, λ2 = 596 nm

Aperture or width of single slit, d = 2 × 10^-6 m

Distance between the slit and the screen, D = 1.5 m

Seperation between the first secondary maximum in the two cases is,

$x2 - x1 = \frac{3}{2} (\frac{Dλ2}{d}) - \frac{3}{2} ( \frac{Dλ1}{d} ) \\ \\ x2 - x1 = \frac{3D}{2d} (λ2 - λ1)$

$\frac{3 \times 1.5}{2 \times 2 \times {10}^{ - 6} } (596 \times {10}^{ - 9} - 590 \times {10}^{ - 9} )$

$\frac{3 \times 1.5 \times 6 \times {10}^{ - 3} }{4} \\ \\ 6.75 \times {10}^{ - 3} m \\ \\ 6.75 \: mm$