JEE Physics Wave Optics MCQs Set 05

Physics Objective Questions and Answers: Wave Optics

Explore reliable objective questions for Wave Optics tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: If the sodium light in Young's double slit experiment is replaced by red light, the fringe width will
(a) Decrease
(b) Increase
(c) Remain unaffected
(d) First increase, then decrease
Show Answer & Explanation

Answer: (b) Increase

Explanation:
1. Red light has a longer wavelength than sodium yellow.
2. Fringe width is proportional to wavelength, so it increases.

Teacher's Note:
1. Longer wavelength gives wider fringes.

Question 2: A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 μm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
(a) Remain unshifted
(b) Shift downward by nearly two fringes
(c) Shift upward by nearly two fringes
(d) Shift downward by 10 fringes
Show Answer & Explanation

Answer: (c) Shift upward by nearly two fringes

Explanation:
1. Extra path \( =(\mu-1)t=0.5\times2\ \mu m=1\ \mu m=2\lambda \).
2. The pattern shifts by two fringes towards the side of the film, which is upward.

Teacher's Note:
1. Central maximum moves towards the slit with the film.

Question 3: In YDSE, when two light waves make the third minimum, they have
(a) A phase difference of 3π
(b) A phase difference of 5π/2
(c) A path difference of 3λ
(d) A path difference of 5λ/2
Show Answer & Explanation

Answer: (d) A path difference of 5λ/2

Explanation:
1. The third minimum corresponds to path difference \( (2\times3-1)\lambda/2=5\lambda/2 \).
2. The phase difference is \( 5\pi \).

Teacher's Note:
1. Nth minimum: (2n - 1) lambda / 2.

Question 4: In a Young's double slit experiment the width between the slits is 3 cm, the separation between slits and screen is 7 cm and the wavelength of light is 1000 Å. The fringe width is
(a) \( 2\times10^{-5}\ \text{m} \)
(b) \( 2\times10^{-9}\ \text{m} \)
(c) \( 0.2\times10^{-6}\ \text{m} \)
(d) \( 2.3\times10^{-7}\ \text{m} \)
Show Answer & Explanation

Answer: (d) \( 2.3\times10^{-7}\ \text{m} \)

Explanation:
1. \( \beta=\lambda D/d=10^{-7}\times0.07/0.03 \).
2. \( \beta=2.3\times10^{-7} \) m.

Teacher's Note:
1. Plug SI values carefully.

Question 5: A monochromatic beam of light is used for forming fringes by illuminating two slits in Young's experiment. When a thin film of mica is interposed in the path of one of the interfering beams then
(a) The fringe width increases
(b) The fringe width decreases
(c) The fringe width remains the same but the pattern shifts
(d) The fringe pattern disappears
Show Answer & Explanation

Answer: (c) The fringe width remains the same but the pattern shifts

Explanation:
1. Mica only changes the optical path.
2. It shifts the pattern but leaves the fringe width unchanged.

Teacher's Note:
1. Same fringe width, shifted pattern.

Question 6: In an interference experiment the third bright fringe is obtained at a point on the screen with light of 700 nm. The wavelength needed to obtain the 5th bright fringe at the same point is
(a) 500 nm
(b) 630 nm
(c) 750 nm
(d) 420 nm
Show Answer & Explanation

Answer: (d) 420 nm

Explanation:
1. Path difference \( =3\times700=2100 \) nm.
2. \( 5\lambda'=2100 \), so \( \lambda'=420 \) nm.

Teacher's Note:
1. Same point means same path difference.

Question 7: In a double slit experiment with light of wavelength 5000 Å the fringe width is 1 mm. If light of 6000 Å is used without altering the system, the new fringe width is
(a) 1 mm
(b) 0.5 mm
(c) 1.2 mm
(d) 1.5 mm
Show Answer & Explanation

Answer: (c) 1.2 mm

Explanation:
1. \( \beta\propto\lambda \).
2. \( \beta'=1\times6000/5000=1.2 \) mm.

Teacher's Note:
1. Direct proportion.

Question 8: Monochromatic green light has wavelength \( 5\times10^{-7} \) m. The separation between the slits is 1 mm. The fringe width on a screen at 2 m distance is
(a) 1 mm
(b) 0.5 mm
(c) 2 mm
(d) 0.1 mm
Show Answer & Explanation

Answer: (a) 1 mm

Explanation:
1. \( \beta=\lambda D/d=5\times10^{-7}\times2/10^{-3} \).
2. \( \beta=10^{-3} \) m = 1 mm.

Teacher's Note:
1. Check units.

Question 9: In Young's double slit experiment, with 700 nm light the fringe width is 0.7 mm. With 500 nm light the fringe width is
(a) 0.35 mm
(b) 0.5 mm
(c) 3.5 mm
(d) 5 mm
Show Answer & Explanation

Answer: (b) 0.5 mm

Explanation:
1. \( \beta\propto\lambda \).
2. \( \beta'=0.7\times500/700=0.5 \) mm.

Teacher's Note:
1. Fringe width scales with wavelength.

Question 10: What is the effect on fringe width when the distance between the slits is doubled?
(a) 1/2 times
(b) 2 times
(c) 1/4 times
(d) Unchanged
Show Answer & Explanation

Answer: (a) 1/2 times

Explanation:
1. \( \beta\propto1/d \).
2. Doubling d halves the fringe width.

Teacher's Note:
1. Inverse proportion to d.

Question 11: The conversation going on in a room can be heard by a person outside the room. The reason for it is
(a) Interference of sound
(b) Reflection of sound
(c) Diffraction of sound
(d) Refraction of sound
Show Answer & Explanation

Answer: (c) Diffraction of sound

Explanation:
1. Sound waves have long wavelengths comparable to the size of the door opening.
2. They bend around corners by diffraction.

Teacher's Note:
1. Diffraction is prominent when the obstacle is comparable to the wavelength.

Question 12: The phenomenon of diffraction occurs
(a) Only in case of light and sound waves
(b) For all kinds of waves
(c) For electromagnetic waves and not for matter waves
(d) For light waves but not in case of X-rays
Show Answer & Explanation

Answer: (b) For all kinds of waves

Explanation:
1. Diffraction is a wave property shown by all waves, including matter waves and X-rays.
2. The other statements are false.

Teacher's Note:
1. Electron diffraction proves matter waves.

Question 13: Which of the following rays gives more distinct diffraction?
(a) X-ray
(b) Light ray
(c) \( \gamma\text{-ray} \)
(d) Radio wave
Show Answer & Explanation

Answer: (d) Radio wave

Explanation:
1. Diffraction is most distinct when the wavelength is comparable to the obstacle size.
2. Radio waves have the longest wavelength of these.

Teacher's Note:
1. Longer wavelength means more bending.

Question 14: Light waves do not travel strictly in straight lines. This can be best explained by
(a) Particle nature of light
(b) Diffraction
(c) Interference
(d) Polarisation
Show Answer & Explanation

Answer: (b) Diffraction

Explanation:
1. Light bends around edges and spreads into the geometrical shadow by diffraction.
2. That shows rectilinear propagation is only approximate.

Teacher's Note:
1. Diffraction shows departure from straight line motion.

Question 15: In the diffraction pattern of a single slit, the width of the central fringe compared to the widths of the other fringes is
(a) Equal
(b) Less
(c) A little more
(d) Double
Show Answer & Explanation

Answer: (d) Double

Explanation:
1. The central maximum extends between the first minima, a width of \( 2\lambda D/a \).
2. Other fringes have width \( \lambda D/a \).

Teacher's Note:
1. Central fringe is twice as wide.

Question 16: The central fringe obtained in the diffraction pattern due to a single slit
(a) Is of minimum intensity
(b) Is of maximum intensity
(c) Has intensity independent of the slit width
(d) None of the above
Show Answer & Explanation

Answer: (b) Is of maximum intensity

Explanation:
1. All the secondary wavelets are in phase at the centre, so the central maximum is brightest.
2. Its intensity depends on the slit width.

Teacher's Note:
1. Central maximum is brightest.

Question 17: In a single slit diffraction pattern, if the light source is replaced by one with a smaller wavelength, the width of the central fringe will be
(a) Less
(b) Increase
(c) Unchanged
(d) None of the above
Show Answer & Explanation

Answer: (a) Less

Explanation:
1. Central width \( =2\lambda D/a \) is proportional to \( \lambda \).
2. A smaller wavelength gives a narrower central fringe.

Teacher's Note:
1. Width is proportional to lambda.

Question 18: In a laboratory, diffraction of light by a single slit is observed. If the slit is made slightly narrower, the diffraction pattern will
(a) Be more spread than before
(b) Be less spread than before
(c) Be spread as before
(d) Disappear
Show Answer & Explanation

Answer: (a) Be more spread than before

Explanation:
1. Width is proportional to \( 1/a \).
2. Narrowing the slit spreads the pattern more.

Teacher's Note:
1. Narrower slit gives wider diffraction.

Question 19: For Fraunhofer single slit diffraction
(a) The width of the central maximum is proportional to λ
(b) On increasing the slit width, the width of the central maximum decreases
(c) On making the slit width a = λ, the central fringe spreads in the range ±90°
(d) All of the above are correct
Show Answer & Explanation

Answer: (d) All of the above are correct

Explanation:
1. Central width \( =2\lambda D/a \) is proportional to \( \lambda \) and decreases with a.
2. For \( a=\lambda \), \( \sin\theta=1 \) so \( \theta=\pm90^\circ \).

Teacher's Note:
1. All three statements follow from a sin(theta) = lambda.

Question 20: In a Fraunhofer diffraction by a slit of width a with wavelength λ and lens focal length f, the linear width of the central maximum is
(a) \( \dfrac{f\lambda}{a} \)
(b) \( \dfrac{fa}{\lambda} \)
(c) \( \dfrac{2f\lambda}{a} \)
(d) \( \dfrac{f\lambda}{2a} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{2f\lambda}{a} \)

Explanation:
1. First minimum at \( y=f\lambda/a \) on either side.
2. The width is \( 2f\lambda/a \).

Teacher's Note:
1. Central maximum is between the first two minima.

Question 21: In a Fraunhofer diffraction obtained by a single slit, the path difference for the nth order minimum is
(a) \( n\lambda \)
(b) \( 2n\lambda \)
(c) \( (2n-1)\lambda/2 \)
(d) \( (2n-1)\lambda \)
Show Answer & Explanation

Answer: (a) \( n\lambda \)

Explanation:
1. Minima occur at \( a\sin\theta=n\lambda \).
2. The path difference is \( n\lambda \).

Teacher's Note:
1. Single slit minima: integer multiples of lambda.

Question 22: A light source of wavelength 5000 Å produces single slit diffraction. The first minimum is at 5 mm from the central maximum and the screen is 2 m from the slit. The slit width is (in mm)
(a) 0.1
(b) 0.4
(c) 0.2
(d) 2
Show Answer & Explanation

Answer: (c) 0.2

Explanation:
1. \( y=\lambda D/a \) gives \( a=\lambda D/y=5\times10^{-7}\times2/(5\times10^{-3}) \).
2. \( a=2\times10^{-4} \) m = 0.2 mm.

Teacher's Note:
1. First minimum at y = lambda D / a.

Question 23: A plane wavefront of wavelength 6000 Å is incident on a slit of width 0.2 mm. The Fraunhofer pattern is obtained on a screen 2 m away. The width of the central maximum is (in mm)
(a) 10
(b) 12
(c) 8
(d) 2
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. Width \( =2\lambda D/a=2\times6\times10^{-7}\times2/(2\times10^{-4}) \).
2. \( =1.2\times10^{-2} \) m = 12 mm.

Teacher's Note:
1. Twice lambda D over a.

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