Physics MCQs for NEET — Practice Questions with Answers

(b) It is given that $\frac{l_2}{l_1}=n\Rightarrow l_2=n{l}_1$

 $\therefore$Ratio of intensites is given by

 $\frac{l_{max}-l_{min}}{l_{max}+l_{min}}=\frac{{\displaystyle {\left(\sqrt{l_2}+\sqrt{l_1}\right)}^2\_{\left(\sqrt{l_2}-\sqrt{l_1}\right)}^2}}{{\left(\sqrt{l_1}+\sqrt{l_2}\right)}^2{\displaystyle +}{\displaystyle {\left(\sqrt{l_2}-\sqrt{l_1}\right)}^2}}$

          $=\frac{{\displaystyle {\left(\sqrt{{\displaystyle \frac{l_2}{l_1}}}{\displaystyle +}{\displaystyle 1}\right)}^2}{\displaystyle -}{\displaystyle {{\displaystyle \left(\sqrt{\frac{{\displaystyle l_2}}{{\displaystyle l_1}}-1}\right)}}^2}}{{\left(\sqrt{\frac{l_2}{l_1}}+1\right)}^2{\displaystyle +}{\displaystyle {{\displaystyle \left(\sqrt{\frac{{\displaystyle l_2}}{{\displaystyle l_1}}-1}\right)}}^2}}$

$=\frac{{\left(\sqrt{n}+1\right)}^2-{\left(\sqrt{n}-1\right)}^2}{{\left(\sqrt{n}+1\right)}^2+{\left(\sqrt{n}-1\right)}^2}=\frac{2\sqrt{n}}{n+1}$

(d) Ist minima is formed at a distance 

                 $Y=\frac{\lambda D}{a}$

For the distance of the first dark band of the diffraction pattern from the centre of the screen is given by position of Ist minima.

 i.e.         $Y=\frac{\lambda D}{a}$

where, $\lambda$=wavelength of parallel beams

          D= focal length

          a= width of linear aperture.

$\Rightarrow y=\frac{\left(5\times {10}^{-5}\right)\left(0.6\right)}{0.02\times {10}^{-2}}$   given

$Y=0.15 cm$

In Young's double-slit experiment, the intensity at the maxima on the screen is given by (2Io), where Io is the intensity from a single slit. When the distance between the screen and the slits is increased, the intensity from each slit decreases due to the spreading of the wavefront. The intensity from a single slit on the screen is inversely proportional to the square of the distance. Therefore, at a distance D = 10d, the intensity from each slit is (Io/10^2) = Io/100, and the total intensity on the screen is (2 * Io/100) = Io/50.

(c) As the first minimum is observed at an angle of ${30}^{\circ }$ in a diffraction pattern due to a single slit of width a. 

i.e.,              $n=1, \theta ={30}^{\circ }$

$\because$ According to bragg's law of diffraction,

       $$\begin{gathered} a \sin \theta =n\lambda \\ \Rightarrow a \sin {30}^{\circ }=\left(1\right)\lambda \left(n=1\right) \\ \Rightarrow a=2\lambda ...\left(i\right) \left\{\because \sin {30}^{\circ }=\frac{1}{2}\right\} \end{gathered}$$

For Ist secondary maxima

$$\begin{gathered} \Rightarrow a \sin {\theta }_1=\frac{3\lambda }{2} \\ \Rightarrow \sin {\theta }_1=\frac{3\lambda }{2a} ...\left(ii\right) \end{gathered}$$

Substitute value of a from eq. (i) to eq (ii), we get

   $$\begin{gathered} \sin {\theta }_1=\frac{3\lambda }{4\lambda }\Rightarrow \sin {\theta }_1=\frac{3}{4} \\ \Rightarrow {\theta }_1={\sin }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$

Given d=1mm=1x10^-3m

D=1m λ=500mm=5x10^-7 m

As width of central maxima=width of 10 maxima

2Dλ/a=10(λD/d)

=> a=d/5=10^-3/5

=0.2x10^-3 m

=0.2mm

At the first minimum adjacent to the central maximum in a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is π radians. This phase difference leads to destructive interference, resulting in the formation of the first minimum.

For net intensity 

I'=4I_o cos² φ/2 (φ=2π/λxλ)

For the first case,

K=4I_o cos2 [π] K=4I_o ...(i)

For the second case

K'=4I_o cos² (π/2/2) (φ=2π/λxλ/4)

=4I_o cos² (π/2)

K'=2I_o ...(ii)

Comparing Eqs. (i)and(ii)

K'=K/2

(c)

$$\begin{gathered} \mathrm{As}, \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}\Rightarrow \mathrm{\lambda }\propto \frac{1}{\mathrm{v}} \\ \mathrm{Therefore}, \mathrm{as} \mathrm{speed} \mathrm{of} \mathrm{electron} \mathrm{increases}, \mathrm{de}-\mathrm{broglie} \mathrm{wavelngth} \mathrm{of} \mathrm{electron} \mathrm{decreases}. \\ \mathrm{Angular} \mathrm{width} \mathrm{of} \mathrm{central} \mathrm{maxima}, \mathrm{w}\propto \mathrm{\lambda }\propto \frac{1}{\mathrm{v}} \\ \mathrm{Then}, \mathrm{width} \mathrm{of} \mathrm{central} \mathrm{maxima} \mathrm{decreases} \mathrm{as} \mathrm{speed} \mathrm{of} \mathrm{electron} \mathrm{increases}. \end{gathered}$$