A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 Å and diffraction bands are observed on a screen 0.5 m from the slit.  The distance of the third dark band from the central bright band is :

  • Option 1)

     3 mm

  • Option 2)

     9 mm

  • Option 3)

     4.5 mm

  • Option 4)

     1.5 mm

     

 

Answers (1)

As we learnt in

Fraunhofer Diffraction -

b\sin \theta = n\lambda
 

- wherein

Condition of nth minima.

b= slit width

\theta = angle of deviation

 

 a = 0.1 mm 10-4 cm 

\lambda=6000\AA=6\times 10^{-7}cm

D = 0.5 m

for 3rd dark bond, a sin\theta=3\lambda

sin\theta=\frac{3\lambda}{a}=\frac{x}{D}

\therefore, x=\frac{3\lambda D}{a}

    x=\frac{3\times 6\times 10^{-7}\times0.5}{10^{-4}}

    x =  9mm 

Correct option is 2.

 


Option 1)

 3 mm

This is an incorrect option.

Option 2)

 9 mm

This is the correct option.

Option 3)

 4.5 mm

This is an incorrect option.

Option 4)

 1.5 mm

 

This is an incorrect option.

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