Wave Optics MCQs for NEET — Physics Questions with Answers

$I=\frac{R_2^2}{4}$ given $n_1{b}_1=n_2{b}_2$ ⇒ $1\times 200=n_2\times 25$

$\therefore n_2=8HPZ$

$\therefore {\left(\frac{{\displaystyle R_9}}{{\displaystyle R_2}}\right)}^{2}I$

$={\left(\frac{{\displaystyle R_9}}{{\displaystyle R_8}}\times \frac{{\displaystyle R_8}}{{\displaystyle R_7}}\times \frac{{\displaystyle R_7}}{{\displaystyle R_6}}\times \frac{{\displaystyle R_6}}{{\displaystyle R_5}}\times \frac{{\displaystyle R_5}}{{\displaystyle R_4}}\times \frac{{\displaystyle R_4}}{{\displaystyle R_3}}\times \frac{{\displaystyle R_3}}{{\displaystyle R_2}}\times \frac{{\displaystyle R_2}}{{\displaystyle R_2}}\right)}^2$

$={\left(\frac{{\displaystyle R_9}}{{\displaystyle R_2}}\right)}^{2}I$

$I=I_0{\left[\frac{{\displaystyle \sin \alpha }}{{\displaystyle \alpha }}\right]}^2,$ where $\alpha =\frac{\varphi }{2}$

For n^th secondary maxima $d\sin \theta =\left(\frac{2n+1}{2}\right)\lambda$

$\Rightarrow \alpha =\frac{\varphi }{2}=\frac{\pi }{\lambda }\left[d\sin \theta \right]=\left(\frac{{\displaystyle 2n+1}}{{\displaystyle 2}}\right)\pi$

$\therefore I=I_0{\left[\frac{{\displaystyle \sin \left(\frac{{\displaystyle 2n+1}}{{\displaystyle 2}}\right)\pi }}{{\displaystyle \left(\frac{{\displaystyle 2n+1}}{{\displaystyle n}}\right)\pi }}\right]}^2=\frac{I_0}{{\left\{\frac{(2n+1)}{2}\pi \right\}}^2}$

So $I_0:I_1:I_2=I_0:\frac{4}{9{\pi }^2}{I}_0:\frac{4}{25{\pi }^2}{I}_0$

$=1:\frac{4}{9{\pi }^2}:\frac{4}{25{\pi }^2}$

Let nth minima of 400 nm coincides with mth minima of 560 nm then,

$(2n-1)400=(2m-1)560$ ⇒ $\frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}$

$$\begin{gathered} For \frac{2n-1}{2m-1}=\frac{7}{5}; \\ \Rightarrow 2n-1=7\Rightarrow n=4 \\ \Rightarrow 2m-1=5\Rightarrow m=3 \end{gathered}$$

i.e. 4th minima of 400 nm coincides with 3rd minima of 560 nm.

The location of this minima is

$=\frac{7\left(1000\right)(400\times {10}^{-6})}{2\times 0.1}=14mm$

Next, 11th minima of 400 nm will coincide with 8th minima of 560 nm

Location of this minima is;

$=\frac{21\left(1000\right)(400\times {10}^{-6})}{2\times 0.1}=42mm$

∴ Required distance = 28 mm

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{I}}_0}{4}={\mathrm{I}}_0{\cos }^2(∆\mathrm{\varphi }/2) \\ \Rightarrow {\cos }^2(∆\mathrm{\varphi }/2)=\frac{1}{4}\Rightarrow ∆\mathrm{\varphi } = 2\mathrm{\pi }/3 \\ \mathrm{So}, \\ \mathrm{\Delta x} \times (2\mathrm{\pi }/\mathrm{\lambda }) =∆\mathrm{\varphi }= 2\mathrm{\pi }/3 \\ \mathrm{\Delta x}= \mathrm{\lambda }/3 \\ \mathrm{For} \mathrm{angular} \mathrm{position} \mathrm{\theta } \mathrm{on} \mathrm{screen}, \\ \sin \mathrm{\theta } = \frac{\mathrm{\Delta x}}{\mathrm{d}} \\ \sin \mathrm{\theta } = \frac{\mathrm{\lambda }}{3\mathrm{d}} \\ \Rightarrow \theta -\mathrm{si}{n}^{-1}\left(\frac{\lambda }{3d}\right) \end{gathered}$$

Momentum of the electron will increase. So the wavelength ($\lambda$ = h/p) of electrons will decrease and fringe width decreases as β ∝ λ.

$I\propto \frac{1}{r^2}$ ⇒ $\frac{\Delta I}{I}=-2\frac{\Delta r}{r}$ = $-2\times 2=-4\%$

Hence intensity is decreased by 4%.

As per the NCERT text, 'A locus of points, which oscillate in phase is called a wavefront; thus a wavefront is defined as a surface of constant phase.' Option o3 directly matches this definition.

The NCERT states, 'Thus, if we wish to determine the shape of the wavefront at t = t, we draw spheres of radius vt from each point on the spherical wavefront where v represents the speed of the waves in the medium. If we now draw a common tangent to all these spheres, we obtain the new position of the wavefront at t = t.' Option o3 accurately describes this process.

The NCERT states, 'Now, according to Huygens principle, each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave.' Option o2 aligns with this.