CBSE Class 12 Physics VBQs Wave Optics

Question. State essential condition for diffraction of light to take place.
Answer : Size of slit is comparable to the size of wavelength.

Question. The angle between the pass axes of a polarizer and analyser is 45°. Write the ratio of the intensitiesof original light and the transmitted light after passing through the analyser.of original light and the transmitted light after passing through the analyser.
Answer : 1/4

Question. The polarizing angle of medium is 600. What is the refractive index of the medium?
Answer : μ = tan ip = tan 600 = √3 = 1.732

Question. At what angle of incidence should a light beam strike a glass slab of refractive index √3, such that the reflected and refracted rays are perpendicular to each other?
Answer : ip =tan -1μ = tan-1 (√3) = 600

Question. The refractive index of a material is 1/ √3. What is the angle of refraction if the unpolarised light is incident on it at the polarizing angle of the medium?
Does the poplarising angle for any transparent medium depend on the wavelength of light?
Answer : ip = tan-1 μ = tan-1(1/√3) = 300
, ip + r = 900 therefore r = 600

Question. Does the poplarising angle for any transparent medium depend on the wavelength of light?
Answer : Yes, since ip = tan−1μ = tan−1(λa⁄λm) since μ = λa⁄λm as refractive index depends upon wavelength.

Question. What type of wavefront will emerge from a (i) point source, and (ii) distant light source?
Answer. (i) Point source – Spherical wavefront (ii) Distant light source – Plane wavefront.

Question. In a single slit diffraction experiment, the width of the slit is reduced to half its original width. How would this affect the size and intensity of the central maximum?
Answer. Β=λD/d=2β ,when d=d/2 Iα(width)2 Hence, I=I/4

Question. Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double slit experiment.
Answer. Two monochromatic sources, which produce light waves, having a constant Phase difference are defined as coherent sources.

Question. Define the term ‘wavefront’
Answer. The wavefront is defined as the locus of all particles of a medium, which are vibrating in the same phase.

Question. Two independent monochromatic sources of light cannot produce a sustained interference pattern. Give reason.
Answer. The phase difference between the light waves originating from two on dependent monochromatic sources will change rapidly with time. The two sources will not be coherent and therefore will not produce a sustained interference pattern.

TWO MARKS QUESTIONS

Question. Why clouds are white?
Answer : The clouds are at much lower heights. They are seen due to scattering of light from lower parts of the atmosphere, which contains large dust particles, water droplets, ice particles etc. all colours scattered equally merge to give us the sensation of white. Hence, clouds appear white.

Question. In Young’s double slit experiment, how is the fringe width altered if the separation between the slits is doubled and the distance between the slits and the screen is halved?
Answer : The fringe width is given by the relation β = λD/d,
Where λ, D and d are the wavelength of light, distance of the screen from the slits and the separation between the slits respectively.
Now, according to the question, d1 =2d and D1 =D/2
Therefore β1 =λD1/d 1 = (λ/2 × 2d) × (β/4) = ( 1/ 4)
Hence, the fringe width decreases to of its value.

Question. A narrow monochromatic beam of light of intensity I is incident a glass plate. Another identical glass plate is kept close to the first one and parallel to it. Each plate reflects 25% of the incident light and transmits the reaming. Calculate the ratio of minimum and maximum intensity in the interference pattern formed by the two beams obtained after reflection from each plate.
Answer : Let I be the intensity of beam I incident on first glass plate.

Each plate reflects 25% of light incident on it and transmits 75%.
Therefore, I1 =I; and I2 = 25/100I = I/4; I3 =75/100 I = 3/4I; I4 = 25/100 I3 = 1⁄4 x 3⁄4 I = 3/16 I
I5= 7/100 I4= 3⁄4 x 3/16 I = 9/64 I
Amplitude ratio of beams 2 and 5 is R = √ I2/I5 = √I/4 x 64/91 = 4/3 Imin/ Imax = [r-1/r+1]2 = [4/3-1 /4/3+1]2 = 1/49 = 1:49

Question. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance D from the slits. If the screen is moved 5 x 10-2 m towards the slits, the charge in fringe width is 3 x 10 -5 m. If the distance between the slit is 10-3 m. Calculate the wavelength of the light used.
Answer : The fringe width in the two cases will be β = Dλ/d; β ‘= D’λ/d
β - β’ = (D-D’)λ/d;
λ = (β - β’ )d / (D-D’)
But D-D’ = 5 x 10-2 m β - β’ = 3 x 10-5 m , d= 10-3 m;
λ = 3 x 10-5 x 10-3 / 5 x 10-2 = 6 x 10-7m= 6000A

Question. A screen is placed 50m from a single slit, which is illuminated with 6000Å light. If the distance between the first and third minima in the diffraction pattern is 3.00mm, what is the width of the slit?
Answer : Here, D=50 cm = 1⁄2 m, λ =6000Å=6×10-7 m
X=3.00 mm = 3×10-3 m, a=?
For the diffraction minima , a sinθ =nλ, when θ is small
Or ax/D=n λ . . . x=n λ D/a
Now x= x3 -x1 = 3 λ D/a-1λD/a=2λD/a . . . a =2 λ D/x=2×6×10-7 ×1/2/3×10-3 =2×10-4 m

Question. Two boys, on their way from school were discussing seriously about something. They both were blowing soap bubbles and were thrilled to watch the expanding bubble with spectacular colour rings. Shwetha, a class XII student, was watching them for a long time, walking behind them. Suddenly she realised that the kids did not look at the traffic in that junction area. She rushed to them and instructed them to be cautious while on the road. She also explained the importance of traffic rules and told them that obeying traffic rules not only makes us safe but also others safe.
a. What are the values highlighted by Shwetha?
b. Why are colours formed on bubbles?
Answers:
a. obeying road rules, alertness, concern in others’s life, clarity of knowledge.
b. due to superposition of incident and reflected waves of white light by thin film(interference)

Question. Ram and Suresh were going to their friend’s house by walk. It was a sunny day in the afternoon. It was very hot. Ram was finding it very difficult to see around him. He had to strain his eyes to see. Suddenly, Suresh took his cooling glasses from his pocket and asked him to wear them and later, Ram slowly managed to see. Suresh advised Ram on the necessity of wearing sun glasses during summer season.
a. What are the values shown by Suresh?
b. Name the phenomenon based on which cooling glasses reduce the glare.
c. What is the resultant intensity of light if both polariser and analyser are rotated through same angle?
Answers:
a. caring , sharing, concern
b. polarisation
c. no change in intensity of light

Question. Jimmy and Johnny were both creating a series of circular waves by jiggling their legs in water. The waves form a pattern. Their friend, Anitha , advised Jimmy and Johnny not to play with water for a long time. She then observed beautiful patterns of ripples which became very colourful. When her friend Latha poured an oil drop on it. Latha, a 12th standard girl, had explained the cause for colourful ripple patterns to Anitha earlier.
a. Identify any 2 values that could be related with Anitha and Latha?
b. Name the phenomenon involved in the above activity?
c. define wavefront
Answers:
a Caring and affectionate
b.Latha is creative and knowledgble
c.Definition of wavefront.

Question. Somya was watching her mother who was washing clothes. Coloured soap bubbles exciting Somya. She asked her mother as to why the soap bubbles appear coloured. Her mother did not know anything about this. In the evening, she talked with Somya’s father. He also did not know anything. Both father and mother went to a science teacher who lived nearby. The teacher explained everything to Somya’s parents. They were satisfied. They came home and explain everything to Somya. Somya was very happy and she expressed her gratitude to her parents.
(a) What according to you are the values displayed by Somya’s parents?
(b) Which concept of physics is involved in the formation of colures in soap bubbles?
Answers:
(a) Somya’s parents showed responsibility towards their daughter. They showed initiative in solving the problem.
(b) Interference.

Question. Rupesh often thought as to how the glare is reduced in headlights of automobiles he once talked to his maternal uncle Rajesh about this. Rajesh himself did not know anything about this. Rajesh went to a library and consulted many books. He could finally understand things. When he explained to Rupesh the details, Rupesh was very happy. Rupesh thanked his uncle Rajesh.
(a) What according to you, are the values displayed by Rajesh?
(b) How glare is reduced in headlights of automobiles?
Answers
(a) Rajesh loves his nephew and is ready to “run the extra mile” for his nephew. Rajesh also showed strong will to find out the details.
(b) Polaroids are used in headlights of automobiles to reduced glare.

Question. Kavita accompanied her mother an optician to buy sunglasses. The shopkeeper showed them a vast variety of sunglasses. The price varied from a few hundred to few thousands. The shopkeeper could not satisfactorily explain the vast difference in the prices of various sunglasses. Kavita’s mother postponed the decision to buy sunglasses. Before buying sunglasses, Kavita’s mother wanted to be sure that she gets the best for Kavita. She went to different shops talked to many peoples and finally bought Polaroid sunglasses for Kavita. Kavita was very happy and she thanked her mother for the best buy.
(a) What according to you are the values displayed by Kavita’s mother?
(b) What is the best about Polaroid sunglasses?
Answers:
(a) Kavita’s mother showed investigative skills. She also showed “perfectionist” attitude. She is a hardworking woman. She is very careful in taking decision.
(b) Polaroid sunglasses drastically red

Question. Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light which produces interference fringes separated by 8.1 mm using same arrangement. Also find the minimum value
of the order ‘n’ of bright fringe of shorter wavelength which coincides with that of the longer wavelength.
Answer. Distance between two bright fringes = Fringe width β=λD/d ,for same values of D and d,we get β₁/β₂=λ₁/λ₂ Substituting and solving we get ,λ₂=720 nm Calculation of minimum value of order for n to be minimum (n + 1)th maxima of shorter wavelength should coincide with nth maximum of longer wavelength (n+1)640= n(720),n=8.Hence minimum order of shorter wavelength is (n+1)=8+1=9

Question. Define the term wavefront. State Huygen’s principle.
Answer. Wave front: Wavefront is locus of all points in which light waves are in same phase. Huygens’ Principle: Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions. These travel with the same velocity as that of the original wavefront.

Question. Two slits are made 1mm apart and the screen is placed away. What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern.
Answer. As per question, width of central maxima of single slit pattern = width of10 maxima of double slit pattern 2λD/a=10(λD/d) a=0.2d , 0.2×10⁻³=0.2×10⁻³m=0.2mm

Question. Yellow light (λ = 6000Å) illuminates a single slit of width 1 x 10^-4 m. Calculate (i)the distance between the two dark lines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit; (ii)the angular spread of the first diffraction minimum.
Answer. (i) Distance between two dark lines, on either side of central maximum is =2λD/a,
Substituting and solving we get,18mm
(ii)Angular spread of the first diffraction minimum (on either side) ϴ=λ/a
Substituting and solving we get ϴ=6x10^-3 rad

Question. A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
Answer. Path difference for nth minimum is asinϴ=nλ or aϴ=nλ
Also,ϴ=x/D From this we get, ax/D=nλ Substituting and solving we get. a=2x10^-4 m

CASE BASED QUESTIONS

1) Refraction of a plane wave

Question. What is the angle made by the ray of light on the wave front?
a) 90˚
b) 0˚
c) 45˚
d) None of the above
Answer. A

Question. Which parameter remains unchanged while a ray of light propagates from one medium to another?
a) velocity
b) Wave length
c) frequency
d) None of the above
Answer. C

Question. According to the above given fig., identify the correct expression for Snell’s law.
a) n1 sin i = n2 sin r
b) n2 sin i = n1 sin r
c) n21 = sin r/ sin i
d) None of the above
Answer. A

Question. When a ray of light travels from a denser to a rarer medium, it
a) it bends towards the normal
b) it travels in a straight line irrespective of angle of incidence.
c) it bends away from the normal
d) None of the above
Answer. C

Question. Which parameter changes while a ray of light propagates from one medium to another?
a) velocity
b) Wave length
c) frequency
d) both a and b
Answer. D

2) Diffraction at a single slit

Question. In the phenomena of Diffraction of light when the violet light is used in the experiment is used instead of red light then,
(a) Fringe width increases
(b) No change in fridge width
(c) Fringe width decreases
(d) Colour pattern is formed
Answer. C

Question. Diffraction aspect is easier to notice in case of the sound waves then in case of the light waves because sound waves
(a) Have longer wavelength
(b) Shorter wavelength
(c) Longitudinal wave
(d) Transverse waves
Answer. A

Question. Diffraction effects show that light does not travel in straight lines. Under what condition the concepts of ray optics are vali(d) ( D = distance of screen from the slit).
(a) D < Zf
(b) D = Zf
(c) D > Zf
(d) D << Zf
Answer. D

Question. when 2nd secondary maxima is obtained in case of single slit diffraction pattern, the angular position is given by
(a) λ
(b) λ/2
(c) 3λ/2
(d) 5λ/2
Answer. D

Question. intensity of secondary maxima in diffraction has
(a) same intensity as central maxima
(b) less intensity than central maxima
(c) more intensity than central maxima
(d) None of these
Answer. C