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:

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

Read more on Brainly.in - https://brainly.in/question/15840403#readmore