Wave Optics Test

Result

Wave Optics Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

An interference is observed due to two coherent sources 'A' & 'B' separated by a distance of 4$$\lambda$$ along the y-axis where $$\lambda$$ is the wavelength of the source. A detector D is moved on the positive x-axis. The number of points on the x-axis excluding the points, x =0 & x =$$\infty$$ at which maximum will be observed is..--

A
three

B
four

C
two

D
infinite

Solution
Question 2
1 / -0

In a young's double slit experiment, d = 1 mm, $$\lambda$$ = 6000 A and D = 1 m (where d, $$\lambda$$ and D have usal meaning). Each of slit individually produces same intensity on the screen. The minimum distance between two points on the screen having 75% intensity of the maximum intensity is:

A
0.45 mm

B
0.40 mm

C
0.30 mm

D
0.20 mm

Solution
Question 3
1 / -0

While an aquarium is being filled with water, a motionless fish looks up vertically through the surface of water at a monochromatic plane wave source of frequency $$f$$. If the index of refraction of the water is $$\mu$$ and the water level rises at a rate $$dh/dt$$, the shift in frequency which the fish observes, will then be :

A
$$\dfrac{(\mu -1)}{c}\dfrac{dh}{dt}$$

B
$$\dfrac{\mu}{c}\dfrac{dh}{dt}$$

C
$$\dfrac{f}{c}.\dfrac{dh}{dt}$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

In a Young's double slit experiment sources of equal intensities are used. Distance between slits is $$d$$ and wavelength of light used is $$\lambda (\lambda <<d)$$, Angular separation of nearest points on either side of central maximum where intensities become half of the maximum value is

A
$$\dfrac{\lambda}{d}$$

B
$$\dfrac{\lambda}{2d}$$

C
$$\dfrac{\lambda}{4d}$$

D
$$\dfrac{\lambda}{6d}$$

Solution
Question 5
1 / -0

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star ; -

A
$$305 \times 10^{-9} $$ radian

B
$$152.5 \times 10^{-9} $$ radian

C
$$610 \times 10^{-9} $$ radian

D
$$457.5 \times 10^{-9} $$ radian

Solution

Limit of resolution of telescope = $$\dfrac{1.22 \lambda}{D}$$
$$\theta = \dfrac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}} = 305 \times 10^{-9}$$ radian
Question 6
1 / -0

Unpolarised light beam of intensity $$I_{0}$$ is incident on polaroid $$P_{1}$$. The three polaroids are arranged in such a way that transmission axis of $$P_{1}$$ and $$P_{3}$$ are perpendicular to each other. Angle between the transmission axis of $$P_{2}$$ and $$P_{3}$$ is $$60^{\circ}$$. The intensity of the beam coming out from $$P_{3}$$ will be

A
$$\dfrac {I_{0}}{2}$$

B
$$\dfrac {3I_{0}}{8}$$

C
$$\dfrac {3I_{0}}{32}$$

D
$$\dfrac {3I_{0}}{64}$$

Solution

$$I_{2} = I_{1}\cos^{2} 30^{\circ} = \dfrac {I_{0}}{2}\times \dfrac {3}{4} = \dfrac {3I_{0}}{8}$$
$$\therefore I_{3} = I_{2}\cos^{2} 60^{\circ} = \dfrac {3I_{0}}{8}\times \dfrac {1}{4} = \dfrac {3I_{0}}{32}$$.
Question 7
1 / -0

The near point of a long-sighted person is $$50\ cm$$ from the eye. Where can she see an object clearly:

A
A distance of $$20\ cm$$

B
Infinity

C
Both $$A$$ and $$B$$

D
None of these

Solution
Question 8
1 / -0

In a Young's double slit experiment the slits are illuminated by a parallel beam of light from the medium of refractive index $$n_{1} = 1.2\ A$$ thin transparent film of thickness $$1.2\mu m$$ and refractive index $$n = 1.5$$ is placed infront of $$s_{1}$$ perpendicular to path of light. Wavelength of light measured in medium $$n_{1}$$ is $$400\ nm$$. The refractive index of medium between plane of slits and screen is $$n_{2} = 1.4$$. If the light coming from the film and $$s_{2}$$ have equal intensities $$I$$ then intensity at geometrical centre of the screen is :

A
$$0$$

B
$$2I$$

C
$$4I$$

D
None of these

Solution
Question 9
1 / -0

In $$YDSE$$ experiment shown in the figure, a parallel beam of light of wavelength $$(\lambda = 0.3\ mm)$$ in medium $$\mu_{1}$$ is incident at an angle $$\theta = 30^{\circ}$$ as shown$$(S_{1}O = S_{2}O)$$. If the intensity due to each light wave at point $$O$$ is $$I_{0}$$ then the resultant intensity at point $$O$$ will be:

A
Zero

B
$$2I_{0} [1 + \cos (40 \pi / 9)]$$

C
$$3I_{0}$$

D
$$I_{0}$$

Solution
Question 10
1 / -0

The YDSE apparatus is as shown in figure below. The condition for point $$P$$ to be a dark fringe is $$(\lambda =$$ wavelength of light waves) :

A
$$(l_{1} - l_{3}) + (l_{2} - l_{4}) = n\lambda$$

B
$$(l_{1} - l_{2}) + (l_{3} - l_{4}) = n\lambda$$

C
$$(l_{1} + l_{3}) - (l_{2} + l_{4}) = \dfrac {(2n - 1)\lambda}{2}$$

D
$$(l_{1} + l_{2}) - (l_{2} + l_{3}) = \dfrac {(2n - 1)\lambda}{2}$$

Solution