Wave Optics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be I= p

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- (a) When a decreases w increases. So, size decreases.

(b) Now, light energy is distributed over a small area and intensity?1/Area  is decreasing so intensity increases

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation- We know that wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, wavelength λ < width of the slit.

Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed On the screen will be di

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b, d)

Explanation-Consider the pattern of the intensity shown in the figure

(i) As intensities of all successive minima is zero, hence we can say that two sources S₁ and S₂ are having same intensities.

(ii) As width of the

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- According to question, there is a hole at point P₂. From Huygen's principle, wave will propagates from the sources S₁ and S₂. Each point on the screen will acts as secondary sources of wavelets. Now, there is a

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation- For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength. In a Young's double-slit experiment, when one of the holes is covered

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

Explanation- due to refraction from the glass medium there is a phase change of $\pi$

$?t$ = $\frac{o{p}^{\text{'}\text{'}}}{v}$ = $\frac{\frac{d}{cosr}}{\frac{c}{n}}=\frac{nd}{ccosr}$

According to snells law n= sin $i$ /sinr

Cosr = $\sqrt[]{1-{sin}^{2}r}=\sqrt[]{1-\frac{{sin}^2\theta }{n^2}}$

$?t$ = $\frac{nd}{c({\frac{{1-sin}^2\theta }{n^2})}^{1/2}}=\frac{n^{2}d}{c}$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

Phase difference = $?\varnothing$ = $\frac{2\pi }{T}\times ?t$ =&n

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- ( a)

Explanation- given width of slit is 10⁴10^-10m= 10^-6

Wavelength of sunlight varies from 4000A⁰ to 8000A⁰

As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the ce

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c)

Explanation-Consider the diagram the light beam incident from air to the glass slab at Brewster's angle (ip ). The incident ray is unpolarised and is represented by dot (.).

The reflected light is plane polarised represente

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- T₂P= T₂O+OP= D+x

T₁P= T₁O-OP= D-x

S₁P $\frac{\mathit{\lambda }}{2}$ = $\sqrt[]{{{(S}_1{T}_1)}^2+{\left(P{T}_1\right)}^2}=\sqrt[]{D^2+{(d-x)}^2}$

S₂P= $\sqrt[]{{{(S}_2{T}_2)}^2+{\left(P{T}_2\right)}^2}=\sqrt[]{D^2+{(d+x)}^2}$

The minima will occur when S₂P- S₁P= (2n-1) $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{D^2+{(d+x)}^2}$ - $\sqrt[]{D^2+{(d-x)}^2}$ = $\frac{\mathit{\lambda }}{2}$

$x=D$

[D²+4D²]^1/2-[D²+0]^1/2= $\frac{\mathit{\lambda }}{2}$

[5D²]^1/2- [D²]^1/2= $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{5}$ D-D= $\frac{\mathit{\lambda }}{2}$

$\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}$ we get D= 0.404 $\mathit{\lambda }$