Wave Optics Test

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Wave Optics Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Two slits in Young's experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, $$\displaystyle \frac{I_{max}}{I_{min}}$$ is

A
$$\displaystyle \frac{4}{9}$$

B
$$\displaystyle \frac{9}{4}$$

C
$$\displaystyle \frac{121}{49}$$

D
$$\displaystyle \frac{49}{121}$$

Solution

Intensity is proportional to width of the slit. thus, $$\dfrac{I_1}{I_2}=\dfrac{1}{25}$$
or $$\dfrac{a_1}{a_2}=\sqrt{\dfrac{I_1}{I_2}}=\dfrac{1}{5}$$ or $$a_2=5a_1$$
now, $$\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)}{(a_1-a_2)^2}=\dfrac{(a_1+5a_1)^2}{(a_1-5a_2)^2}=36/16=9/4$$
Question 2
1 / -0

The ratio of resolving powers of an optical microscope for two wavelengths $$\lambda_1=4000\mathring{A}$$ and $$\lambda_2=6000\mathring{A}$$ is:

A
$$16:18$$

B
$$8:27$$

C
$$9:4$$

D
$$3:2$$

Solution

Resolving power $$ \propto \dfrac{1}{\lambda}$$

$$\dfrac{RP_1}{RP_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{6000}{4000}$$

$$\dfrac{RP_1}{RP_2}=\dfrac{3}{2}$$
Question 3
1 / -0

In Youngs double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

A
half

B
four times

C
one-fourth

D
double

Solution

Fringe width $$ \beta =\dfrac { \lambda D }{ d } $$
On $$ d'=\dfrac { d }{ 2 } ,D'=2D$$
New fringe width $$\beta '=\dfrac { \lambda D' }{ d' } =4\beta $$
Question 4
1 / -0

Assume that light of wavelength $$600 nm$$ is coming from a star. The limit of resolution of telescope whose objective has a diameter of $$2m$$ is :

A
$$1.83\times 10^{7} rad$$

B
$$7.32 \times 10^{-7} rad$$

C
$$6.00 \times 10^{7} rad$$

D
$$3.66 \times 10^{-7} rad$$

Solution

Diameter of onjective of telescope,
$$D=2\ m=100×2\ cm=200\ cm$$

The wavelength of light is:
$$\lambda=600\ nm=6\times10^{−5}\ cm$$

Limit of resolution of telescope,
$$d\theta=\dfrac{1.22\lambda}{D}$$

$$=\dfrac{1.22\times6\times10^{−5}}{200}$$

$$=3.66\times10^{−7}$$ radian
Question 5
1 / -0

If $$f_{0} = 5\ cm, \lambda = 600{A^{0}}, a = 1\ cm$$ for a microscope, then what will be its resolving power.

A
$$11.9\times 10^{5} / m$$

B
$$10.9\times 10^{5} / m$$

C
$$10.9\times 10^{4} / m$$

D
$$10.9\times 10^{3} / m$$

Solution

$$R.P. = \dfrac {2\mu \sin \theta}{1.22\lambda}$$
$$\mu = 1$$
$$\tan \theta \simeq \sin \theta = \dfrac {a}{f} = \dfrac {1}{5} = 0.2\Rightarrow R.D. = \dfrac {2\times 1\times 0.2}{1.22\times 6\times 10^{-7}} = \dfrac {4\times 10^{6}}{3.66}$$
$$RP. = 10.9\times 10^{5} / m$$.
Question 6
1 / -0

In a single slit diffraction with $$\displaystyle \lambda =500nm$$ and a lens of diameter 0.1 mm, width of central maxima, obtain on screen at a distance of 1 m will be

A
$$5 mm$$

B
$$1 mm$$

C
$$10 mm$$

D
$$2.5 mm$$

Solution

Angle subtended by two minima at the slit in a single slit diffraction $$=\alpha=\dfrac{2\lambda}{w}$$
where, $$w$$ is the slit width.
Here the lens' diameter would act as slit width.
The width of central maxima is the distance between the two minima $$=d\alpha$$
where, $$d$$ is the distance between slit and screen $$=1\ m$$
Thus, the width of central maxima $$=10\ mm$$
Question 7
1 / -0

In a $$YDSE$$, the light of wavelength $$\lambda = 5000\overset {\circ}{A}$$ is used, which emerges in phase from two slits a distance $$d = 3\times 10^{-7}m$$ apart. A transparent sheet of thickness $$t = 1.5\times 10^{-7}m$$ refractive index $$\mu = 1.17$$ is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of $$y$$.

A
$$4.9^{\circ}$$ and $$\dfrac {D(\mu - 1)t}{2d}$$

B
$$4.9^{\circ}$$ and $$\dfrac {D(\mu - 1)t}{d}$$

C
$$3.9^{\circ}$$ and $$\dfrac {D(\mu + 1)t}{d}$$

D
$$2.9^{\circ}$$ and $$\dfrac {D(\mu + 1)t}{d}$$

Solution

The path difference when transparent sheet is introduced $$\triangle x = (\mu - 1)t$$
If the central maxima occupies position of $$nth$$ fringe, then $$(\mu - 1) t = n\lambda = d\sin \theta$$
$$\Rightarrow \sin \theta = \dfrac {(\mu - 1)t}{d} = \dfrac {(1.71 - 1)\times 1.5\times 10^{-7}}{3\times 10^{-7}} = 0.085$$
Therefore, angular position of central maxima
$$\theta = \sin^{-1} (0.085) = 4.88^{\circ} \approx 4.9$$
For small angles, $$\sin \theta \approx \theta \approx \tan \theta$$
$$\Rightarrow \tan \theta = \dfrac {y}{D}$$
$$\therefore \dfrac {y}{D} = \dfrac {(\mu - 1)t}{d}\Rightarrow y = \dfrac {D(\mu - 1)t}{d}$$.
Question 8
1 / -0

A : The phase difference between any two points on a wave front is zero
R : From the source light, reaches every point on the wave front in the same time.

A
Both A and R are true, and R is correct explanation of A

B
Both A and R are true, and R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

A wave front is the locus of points having the same phase.
Thus statement A is true because the phase difference is zero as all points have same phase.
Statement R is true because a wavefront is composed of all points where the light reaches from the source in the same time.
As light takes the same time it has the same phase at all points on a wavefront and hence it correctly explains statement A.
Question 9
1 / -0

A plane wave front falls on a convex lens. The emergent wave front:

A
converges to a point

B
diverges from a point

C
does not suffer any refraction

D
may or may not converge at point

Solution

Although a convex lens is s converging lens, it does not ensure that all emerging wave front would be converging. It depends upon the incident wave front. It only has a converging tendency. There may be cases when a diverging wave front is not converged enough, and the emergent wave front is still diverging.
However a plane wave front would always be converged by a convex lens.
Question 10
1 / -0

Huygens' wave theory is used :

A
to determine the velocity of light

B
to find the position of the wave front

C
to determine the wavelength of light

D
to find the focal length of a lens

Solution

Huygen proposed a hypothesis for the geometrical construction of the position of a common wavefront at any instant during the propogation of waves in a medium.

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Re-Attempt Weekly Quiz Competition

Wave Optics Test - 13

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Wave Optics Test - 13

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SHARING IS CARING

Question 1

$$\displaystyle \frac{4}{9}$$

$$\displaystyle \frac{9}{4}$$

$$\displaystyle \frac{121}{49}$$

$$\displaystyle \frac{49}{121}$$

Solution

Question 2

$$16:18$$

$$8:27$$

$$9:4$$

$$3:2$$

Solution

Question 3

half

four times

one-fourth

double

Solution

Question 4

$$1.83\times 10^{7} rad$$

$$7.32 \times 10^{-7} rad$$

$$6.00 \times 10^{7} rad$$

$$3.66 \times 10^{-7} rad$$

Solution

Question 5

$$11.9\times 10^{5} / m$$

$$10.9\times 10^{5} / m$$

$$10.9\times 10^{4} / m$$

$$10.9\times 10^{3} / m$$

Solution

Question 6

$$5 mm$$

$$1 mm$$

$$10 mm$$

$$2.5 mm$$

Solution

Question 7

$$4.9^{\circ}$$ and $$\dfrac {D(\mu - 1)t}{2d}$$

$$4.9^{\circ}$$ and $$\dfrac {D(\mu - 1)t}{d}$$

$$3.9^{\circ}$$ and $$\dfrac {D(\mu + 1)t}{d}$$

$$2.9^{\circ}$$ and $$\dfrac {D(\mu + 1)t}{d}$$

Solution

Question 8

Both A and R are true, and R is correct explanation of A

Both A and R are true, and R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 9

converges to a point

diverges from a point

does not suffer any refraction

may or may not converge at point

Solution

Question 10

to determine the velocity of light

to find the position of the wave front

to determine the wavelength of light

to find the focal length of a lens

Solution

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