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Wave Optics

NEET Physics · 95 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550 nm . [Assume that the light is incident nearly perpendicular to the glass surface.]
A 68.7 nm
B 137.5 nm
C 94.8 nm
D 275 nm
Correct Answer
Option B
Solution

For transmitted green light to be maxima, reflected green should be minima.

ΔP=2μ0t=nλ⇒t=nλ2μ0∴tmin⁡=λ2μ0=5502×2=137.5\begin{aligned} & \Delta \mathrm{P}=2 \mu_0 \mathrm{t}=\mathrm{n} \lambda \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu_0} \therefore \mathrm{t}_{\min }=\frac{\lambda}{2 \mu_0}=\frac{550}{2 \times 2}=137.5 \end{aligned}
Q52
Two polarisers P1P_1 and P2P_2 are placed in such a way that the intensity of the transmitted light will be zero. A third polariser P3P_3 is inserted in between P1P_1 and P2P_2, at particular angle between P2P_2 and P3P_3. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers P2P_2 and P3P_3 is :
A π/6\pi / 6
B π3\dfrac{\pi}{3}
C π4\dfrac{\pi}{4}
D π/8\pi / 8
Correct Answer
Option C
Solution

Through P2I1=I0sin⁡2(π2−θ)P_2 I_1=I_0 \sin ^2\left(\dfrac{\pi}{2}-\theta\right)

I1=I0cos⁡2θI_1=I_0 \cos ^2 \theta

Through P3Inet =(I0cos⁡2θ)sin⁡2θP_3 I_{\text{net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta

Inet =I04[sin⁡(2θ)]2 for max Inet θ=45∘I_{\text{net }}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text{ for max } I_{\text{net }} \theta=45^{\circ}

So angle between P2\mathrm{P}_2 and P3=π4\mathrm{P}_3=\dfrac{\pi}{4} Correct Ans. (1)

Q53
In a Young's double slit experiment, the slits are separated by 0.2 mm . If the slits separation is increased to 0.4 mm , the percentage change of the fringe width is :
A 25%25 \%
B 50%50 \%
C 0%0 \%
D 100%100 \%
Correct Answer
Option B
Solution

In Young's double slit experiment, the fringe width β\beta is given by:

β=λDd\beta = \frac{\lambda D}{d}

where: λ\lambda is the wavelength of light,

DD

is the distance from the slits to the screen,

dd

is the separation between the slits. Here's how the change affects the fringe width: Initial Situation: When

d=0.2 mmd = 0.2 \text{ mm}

, the fringe width is:

βinitial=λD0.2 mm\beta_{\text{initial}} = \frac{\lambda D}{0.2 \text{ mm}}

After Increasing Slit Separation: When

dd

is increased to

0.4 mm0.4 \text{ mm}

:

βnew=λD0.4 mm\beta_{\text{new}} = \frac{\lambda D}{0.4 \text{ mm}}

Comparing the Two Fringe Widths: Notice that:

βnew=12×λD0.2 mm=12βinitial\beta_{\text{new}} = \frac{1}{2} \times \frac{\lambda D}{0.2 \text{ mm}} = \frac{1}{2} \beta_{\text{initial}}

This means the fringe width is halved, which is a reduction of

50%50\%

. Thus, the fringe width decreases by

50%50\%

when the slit separation is increased from

0.2 mm0.2 \text{ mm}

to

0.4 mm0.4 \text{ mm}

. The correct answer is Option B:

50%50\%

.

Q54
Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year = 9.46 × \times 1015 m) of the order of :
A 106 km
B 108 km
C 1011 km
D 1010 km
Correct Answer
Option B
Solution

The limit of resolution of a telescope,

Δ\Delta

θ\theta =

1.22  λD{{1.22\,\,\lambda } \over D}

=

lR{l \over R}

∴\therefore

ll

=

1.22  λRD{{1.22\,\,\lambda R} \over D}

=

1.22×6×10−7×10×9.46×101530×10−2{{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}

= 2.31 ×\times 108 km

Q55
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ\lambda = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range −- 30o≤ \le θ\theta ≤ \le 30o is :
A 640
B 320
C 321
D 641
Correct Answer
Option D
Solution

We know, path difference, d sinθ\theta = nλ\lambda here n = no of bright fringer in the angle here given d = 0.32 ×\times 10-3 m λ\lambda = 500 ×\times 10−-9 m ∴\therefore 0.32 ×\times 10−-3 sin30o = n ×\times 500 ×\times 10−-9 ⇒\Rightarrow n = 320 Total number of maxima in the range −- 30o θ\theta ≤\le 30o is = 320 ×\times 2 + 1 = 641

Q56
In a double slit experiment, when a thin film of thickness t having refractive index μ\mu . is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is (λ\lambda is the wavelength of the light used) :
A λ2(μ−1){\lambda \over {2\left( {\mu - 1} \right)}}
B λ(2μ−1){\lambda \over {\left( {2\mu - 1} \right)}}
C 2λ(μ−1){{2\lambda } \over {\left( {\mu - 1} \right)}}
D λ(μ−1){\lambda \over {\left( {\mu - 1} \right)}}
Correct Answer
Option D
Solution

As we know, Path difference introduced by thin film,

Δ=(μ−1)t\Delta=(\mu-1) t

.......(i) and if fringe pattern shifts by one frings width, then path difference,

Δ=1×λ=λ\Delta=1 \times \lambda=\lambda

.......(ii) So, from Eqs. (i) and (ii), we get

(μ−1)t=λ(\mu-1) t=\lambda

⇒\Rightarrow

t=λμ−1t=\frac{\lambda}{\mu-1}
Q57
With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 Ao\mathop A\limits^o is observed at 5896 Ao\mathop A\limits^o ?
A 306 km/sec
B 322 km/sec
C 296 km/sec
D 336 km/sec
Correct Answer
Option A
Solution
f=f01+β1−βf = {f_0}\sqrt {{{1 + \beta } \over {1 - \beta }}}
β=vc\beta = {v \over c}
ff0=1+β1−β{f \over {{f_0}}} = \sqrt {{{1 + \beta } \over {1 - \beta }}}
(1+Δff0)2=(1+β)(1−β)−1{\left( {1 + {{\Delta f} \over {{f_0}}}} \right)^2} = (1 + \beta ){(1 - \beta )^{ - 1}}

β\beta is small compared to 1

(1+2Δff0)=(1+2β)\left( {1 + {{2\Delta f} \over {{f_0}}}} \right) = (1 + 2\beta )
β=Δff0=vc\beta = {{\Delta f} \over {{f_0}}} = {v \over c}
v=6×c5890=305.6v = 6 \times {c \over {5890}} = 305.6

km/s

Q58
The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μ\mu m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)
A 100 μ\mu m
B 25 μ\mu m
C 50 μ\mu m
D 75 μ\mu m
Correct Answer
Option B
Solution

Given 2θ\theta = 60o ⇒\Rightarrow θ\theta = 30o We know,

aa

sin θ\theta = n λ\lambda for first minima n = 1, ∴\therefore At first minima

aa

sin = λ\lambda ⇒\Rightarrow

   \,\,\,

10−-6 ×\times sin 30o = λ\lambda ⇒\Rightarrow

   \,\,\,

λ\lambda =

10−62{{{{10}^{ - 6}}} \over 2}

m Now after making a new slit, ∴\therefore Fringe width, β\beta =

λDd{{\lambda D} \over d}

given, D = 50 cm and β\beta = 1 cm. ∴\therefore 1 ×\times 10−-2 =

0.5×10−6×50×10−2d{{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}

⇒\Rightarrow d = 25 ×\times 10−-6 m = 25 μ\mum.

Q59
In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case:
A there shall be alternate interference fringes of red and green.
B there shall be an interference pattern for red distinct from that for green.
C there shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.
D there shall be no interference fringes.
Correct Answer
Option D
Solution

When white light is used in Young's double slit experiment and one slit is covered by a red filter and the other by a green filter, several effects occur: Different Fringe Widths: Each color will produce fringes of different widths due to their varying wavelengths.

This means that red and green will not overlap perfectly.

Color Overlap: As the fringe patterns develop, some regions will have overlap due to the differing fringe spacings of red and green light.

This overlap will create areas where the colors mix, potentially leading to an appearance of intermediate colors.

Overall, because each color produces its own pattern and these do not align, a clear interference pattern as seen with monochromatic light (a single color) will not be formed.

Instead, there will be a complex pattern where the fringes do not distinctly appear as solid bands of red or green due to the overlapping and different fringe widths.

Q60
A light whose electric field vectors are completely removed by using a good polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism.
A Reflected and refracted rays will be perpendicular to each other.
B Wave will propagate along the surface of prism.
C No refraction, and there will be total reflection of light.
D No reflection, and there will be total transmission of light.
Correct Answer
Option D
Solution

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

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