Physics

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 centre all colours appear i.e., mixing of colours form white patch at the centre.

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 represented by arrows.

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

This is a short answer type question as classified in NCERT Exemplar

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 $\mathit{\lambda }$

This is a short answer type question as classified in NCERT Exemplar

I_rms= $\sqrt[]{\frac{1^2+2^2}{2}}$ = $\sqrt[]{\frac{5}{2}}$ = 1.58A = 1.6 approx

This value is indicating in graph

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

Explanation- resolving power =1/d = $\frac{2sin\beta }{1.22\lambda }$  = d_min= $\frac{1.22\lambda }{2sin\beta }$  where $\lambda$  is the wavelength of light

So d_min= $\frac{1.22*5500*{10}^{-10}}{2sin\beta }$

$\lambda$ = $\frac{12.27}{\sqrt[]{V}}$ = 0.12 $*$ 10^-9m

$\frac{d\text{'}min}{dmin}$ = $\frac{0.12*{10}^{-9}}{5500*{10}^{-10}}$ = 0.2 $*$ 10^-3

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

Explanation- When angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented By arrows.

Polarisation by reflection occurs when the angle of incidence Is the Brewster's angle

So tani_b = $\frac{{\mu }_1}{{\mu }_2}$  where $\mu$ ₂< $\mu$ ₁

When the light rays travels in such a medium, the critical angle is

Sini_c= $\frac{{\mu }_2}{{\mu }_1}$

Where $\mu$ ₂< $\mu$ ₁

As tani_b > sini_c for large angles i_bc

Thus the polarisation by reflection occurs definitely.

This is a short answer type question as classified in NCERT Exemplar

In the figure

Band width = w₂-w₁ where these two corresponds to frequencies at which magnitude of current is 1/ $\sqrt[]{2}$ times of maximum value

I_rms= $\frac{I_{max}}{\sqrt[]{2}}$ = 1/ $\sqrt[]{2}$ = 0.7A

From the diagram the corresponding frequencies are 0.8 and 1.2rad/s

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

Explanation- angular resolution of the eye? = 5.8 x 10^-4

linear distance between two dots is l= 2.54/300=0.84 $*$ 10^-2cm

? = l/z

z=l/? = $\frac{0.84*{10}^{-2}}{5.8*{10}^{-4}}$ = 14.5cm