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Question: A linear aperture whose width is \( 0.02{\text{ }}cm \) is placed immediately in front of a lens of ...

A linear aperture whose width is 0.02 cm0.02{\text{ }}cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5×10−5cm5 \times {10^{ - 5}}cm . The distance of the first dark band of the diffraction pattern from the center of the screen is:
A) 0.15 cm0.15\,cm
B) 0.10 cm0.10\,cm
C) 0.25 cm0.25\,cm
D) 0.20 cm0.20\,cm

Explanation

Solution

Hint The focal length of the lens is where the image of the parallel beam will form. The distance of the first dark band of the diffraction pattern is when the path difference is equal to integral multiples of the wavelength.
⇒y=λDd\Rightarrow y = \dfrac{{\lambda D}}{d} where yy is the distance of the first dark band from the centre of the screen that is at a distance DD away from a slit of width dd .

Complete step by step answer
Since the linear aperture is placed in front of the lens of focal length 60 cm, the image of the parallel beam will form on the aperture and a diffraction pattern will form on the screen.
So f=D=60 cmf = D = 60\,cm.
The distance of the first dank band can be calculated using the fact that for a diffraction minimum, the path difference must be equal to integral multiples of the wavelength of light. So, we have
⇒dsin⁡θ=nλ\Rightarrow d\sin \theta = n\lambda
For θ→0\theta \to 0 , we can write sin⁡θ=θ=yD\sin \theta = \theta = \dfrac{y}{D} . On substituting this value for the first minima i.e. n=1n = 1 in the above equation, we get:
⇒ydD=λ\Rightarrow \dfrac{{yd}}{D} = \lambda
On rearranging the terms, we get
⇒y=λDd\Rightarrow y = \dfrac{{\lambda D}}{d}
Substituting the values of λ=5×10−5cm\lambda = 5 \times {10^{ - 5}}cm , D=60 cmD = 60\,cm and d=0.02 cmd = 0.02{\text{ }}cm , we get
⇒y=4×10−5×60 0.02 \Rightarrow y = \dfrac{{4 \times {{10}^{ - 5}} \times 60\,}}{{0.02{\text{ }}}}
⇒ 0.15 cm\Rightarrow \,0.15\,cm
Hence the first minima is located at a distance of 0.15 cm0.15\,cm from the centre of the screen which corresponds to option (A).

Note
The only significance of the lens in the question is that it creates an image exactly on the aperture. The condition for minima and maxima in a diffraction pattern are the opposites of the conditions for an interference pattern and must not be confused with each other.