Wave Optics

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 $\times$ 10^-2cm

? = l/z

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

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

Explanation- As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wa

This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. Due to the large distance the radius of wavefront can be considered a

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

Explanation- The point image I₁, due to L₁ is at the focal point. Now, due to the converging lens L₂, let final image formed is I which is point image, hence the wavefront for this image will be of spherical symmetry.

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

Explanation- When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of

wavefront is in accordance with Huygen's principle.

So,

This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- refractive index = 1.38 refractive index = 1.5

^{$\lambda =5500A$ 0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass i

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

Explanation-All points with the same optical path length must have the same phase.

So – $\sqrt[]{{\mu }_r{\epsilon }_r}AE$   =BC-–  $\sqrt[]{{\mu }_r{\epsilon }_r}CD$

BC= $\sqrt[]{{\mu }_r{\epsilon }_r}$ (CD-AE)

BC>0, si must be greater than AD

But in other figure

–  $\sqrt[]{{\mu }_r{\epsilon }_r}AE=BC–\mathrm{ }\mathrm{ }\sqrt[]{{\mu }_r{\epsilon }_r}CD$

So BC= –  $\sqrt[]{{\mu }_r{\epsilon }_r}\left(CD-AE\right)$

But clearly here BE is less

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

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt)  and path difference is $\frac{\lambda }{2}$  hence phase difference is $\pi$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $\pi$ )= - acoswt here path d

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

Explanation- as the refractive index of the class $\mu$ , the path difference will be calculated as $?x$ =2dsin $\theta$ +( $\mu -1$ )L

For principal maxima ,(path difference is zero)

2dsin $\theta$ ₀+( $\mu -1$ )L=0

Sin $\theta$ ₀= - $\frac{L(\mu -1)}{2d}$ = $\frac{-L\left(0.5\right)}{2d}$

Sin $\theta$ ₀=-1/16

OP=Dtan $\theta$ ₀= Dsin $\theta$ ₀=-D/16

For p

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

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\varnothing$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\varnothing$ )

A_R=a(2cos $\varnothing /2$ sin(wt+ $\varnothing$ ))

A_R=2acos $\varnothing /2$  sin(wt+ $\varnothing$ )

This eqn is also same for