Practice Start Test

Wave Optics

NEET Physics · 95 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index nn) is :
A tan⁡−1(1/n){\tan ^{ - 1}}\left( {1/n} \right)
B sin⁡−1(1/n){\sin ^{ - 1}}\left( {1/n} \right)
C sin⁡−1(n){\sin ^{ - 1}}\left( n \right)
D tan⁡−1(n){\tan ^{ - 1}}\left( n \right)
Correct Answer
Option D
Solution

The angle of incidence for total polarization is given by

tan⁡θ=n⇒θ=tan⁡−1n\tan \theta = n \Rightarrow \theta = {\tan ^{ - 1}}n

Where

nn

is the refractive index of the glass.

Q42
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is :
A three
B five
C infinite
D zero
Correct Answer
Option B
Solution

For constructive interference

d sin⁡θ=nλd\,\sin \theta = n\lambda

Given

d=2λ⇒sin⁡θ=n2d = 2\lambda \Rightarrow \sin \theta = {n \over 2}
n=0,1,−1,2,−2n = 0,1, - 1,2, - 2

hence five maxima are possible.

Q43
Two point white dots are 11 mmmm apart on a black paper. They are viewed by eye of pupil diameter 33 mm.mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light =500=500 nmnm ]
A 1m1m
B 5m5m
C 3m3m
D 6m6m
Correct Answer
Option B
Solution
yD≥1.22λd{y \over D} \ge 1.22{\lambda \over d}
⇒D≤yd(1.22)λ\Rightarrow D \le {{yd} \over {\left( {1.22} \right)\lambda }}
=10−3×3×10−3(1.22)×5×10−7= {{{{10}^{ - 3}} \times 3 \times {{10}^{ - 3}}} \over {\left( {1.22} \right) \times 5 \times {{10}^{ - 7}}}}
=3061≈5m= {{30} \over {61}} \approx 5m

∴\therefore

Dmax⁡=5m{D_{\max }} = 5m
Q44
A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
A circle
B hyperbola
C parabola
D straight line
Correct Answer
Option D
Solution

The interference fringes formed in a Young's double slit experiment are straight lines.

These fringes are a result of the constructive and destructive interference of the light waves coming from the two slits, and they appear as alternating bright and dark straight bands or lines.

Hence, the answer is : Option D : straight line.

Remember for double hole experiment a hyperbola is generated.

Q45
Two beams, AA and BB, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam AA has maximum intensity (and beam BB has zero intensity), a rotation of polaroid through 30∘{30^ \circ } makes the two beams appear equally bright. If the initial intensities of the two beams are IA{{\rm I}_A} and IB{{\rm I}_B} respectively, then IAIB{{{{\rm I}_A}} \over {{{\rm I}_B}}} equals:
A 33
B 32{3 \over 2}
C 11
D 13{1 \over 3}
Correct Answer
Option D
Solution

According to malus law, intensity of emerging beam is given by,

I=I0cos⁡2θI = {I_0}{\cos ^2}\theta

Now,

IA′=IAcos⁡230∘{I_{A'}} = {I_A}{\cos ^2}{30^ \circ }
IB′=IBcos⁡260∘{I_{B'}} = {I_B}{\cos ^2}{60^ \circ }

As

IA′=IB′{I_{A'}} = {I_{B'}}
⇒IA×34=IB×14\Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}

∴\therefore

IAIB=13{{{I_A}} \over {{I_B}}} = {1 \over 3}
Q46
In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be
A 28
B 24
C 30
D 18
Correct Answer
Option A
Solution

Let the length of segment is "

ll

" Let N is the no. of fringes in "

ll

" and w is fringe width. ∴\therefore Nw =

ll

⇒\Rightarrow N

(λDd)\left( {{{\lambda D} \over d}} \right)

=

ll

As in both cases segment length is same. ∴\therefore

N1λ1Dd=l{{{{N_1}{\lambda _1}D} \over d} = l}

and

N2λ2Dd=l{{{{N_2}{\lambda _2}D} \over d} = l}

∴\therefore

N1λ1Dd{{{{N_1}{\lambda _1}D} \over d}}

=

N2λ2Dd{{{{N_2}{\lambda _2}D} \over d}}

⇒\Rightarrow

N1λ1{{N_1}{\lambda _1}}

=

N2λ2{{N_2}{\lambda _2}}

⇒\Rightarrow 16 × 700 = N2 × 400 ⇒\Rightarrow N2 = 28

Q47
In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :
A 0.50 mm
B 0.25 mm
C 1 mm
D 0.75 mm
Correct Answer
Option B
Solution

Fringe width (β\beta) =

λDd{{\lambda D} \over d}

d = 2 ×\times 10−-3 m λ\lambda = 500 ×\times 10−-9 m D = 1 m Now β\beta =

500×10−9×12×10−3{{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}

β\beta =

52×10−4{5 \over 2} \times {10^{ - 4}}

β\beta = 2.5 ×\times 10−-4 β\beta = 0.25 mm

Q48
In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Ao\mathop A\limits^o is :-
A 1178 ×\times 10−-9 m
B 1178 ×\times 10−-6 m
C 1178 ×\times 10−-12 m
D 5890 ×\times 10−-7 m
Correct Answer
Option B
Solution

In the double-slit experiment, the position of a bright fringe is given by the formula : y=mλLdy = \dfrac{{m\lambda L}}{{d}}, where : m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.) λ\lambda is the wavelength of the light (in meters) L is the distance from the slits to the screen (in meters) d is the distance between the slits (in meters) We need to find the difference in position between the first and third bright fringes.

So, we find the position of both and subtract the position of the first from the position of the third : y3=3λLdy_3 = \dfrac{{3\lambda L}}{{d}} y1=λLdy_1 = \dfrac{{\lambda L}}{{d}} Then, the difference between the third and the first bright fringes is : y3−y1=2λLdy_3 - y_1 = 2\dfrac{{\lambda L}}{{d}} Now, let's plug the given values: λ=5890 A˚=5890×10−10 m\lambda = 5890 \,Å = 5890 \times 10^{-10} \, m (since 1 Å = 10−1010^{-10} meters), L = 0.5 m, and d = 0.5 mm = 0.5 ×10−3\times 10^{-3} m :

y3−y1=2×5890×10−10 m×0.5 m0.5×10−3 m=1178×10−6 my_3 - y_1 = 2 \times \frac{5890 \times 10^{-10} \, \text{m} \times 0.5 \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = 1178 \times 10^{-6} \, \text{m}

So, the correct answer is 1178 ×10−6\times 10^{-6} m, which corresponds to Option B.

Q49
Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)
A 3 : 4
B 4 : 3
C 7 : 1
D 49 : 1
Correct Answer
Option D
Solution

Let, Width of first slit = w1 and width of second slit = w2 Given,

w1w2=916{{{w_1}} \over {{w_2}}} = {9 \over {16}}

Given, Intensity of light

(I)∝w(I) \propto w

∴\therefore

I1I2=w1w2=916{{{I_1}} \over {{I_2}}} = {{{w_1}} \over {{w_2}}} = {9 \over {16}}

We know,

Imax⁡Imin⁡=(I1+I2I1−I2)2{{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}
=(I1I2+1I1I2−1)2= {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}
=(34+134−1)2= {\left( {{{{3 \over 4} + 1} \over {{3 \over 4} - 1}}} \right)^2}
=(71)2= {\left( {{7 \over 1}} \right)^2}
=491= {{49} \over 1}
Q50
A beam of unpolarised light of intensity I0I_0 is passed through a polaroid AA and then through another polaroid BB which is oriented so that its principal plane makes an angle of 45∘45^{\circ} relative to that of AA. The intensity of emergent light is:
A I0/2I_0 / 2
B I0/8I_0 / 8
C I0/4I_0 / 4
D I0I_0
Correct Answer
Option C
Solution

Intensity of emergent light

=I02cos⁡245∘=I04=\frac{\mathrm{I}_0}{2} \cos ^2 45^{\circ}=\frac{\mathrm{I}_0}{4}
Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →