Wave Optics MCQ Class XII. (TN State Board Syllabus) pptx

MCQ (Multiple choice Questions) in Wave Optics - Class XII Physics based TN State Board (With Explanation)

1. Wave Optics MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is, red (b) yellow (c) green (d) violet When the plane is placed over various coloured letters, the upward shift is given by, S = t[ 1 – 1/n] ie . S t[ 1 – 1/n] n – Refractive index We know that n (Since λ 1/ λ 2 = n2/n1) From the above relations, we know that the upward shift (raise) is more for a colour with more n and less λ . ie . violet Ans: d  

4. 2. Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm] 1 m (b) 5 m (c) 3 m (d) 6m Angular resolution is θ = x/R = 1.22 λ /a ie . R = ax/1.22 λ R - The maximum distance that can be resolved λ – Wavelength of light = 500nm = 500x10 -9 m = 5x10 -7 m a – Diameter of the pupil = 3mm = 3x10 -3 m x – Distance between the two dots = 1mm = 1x10 -3 m

5. Contd … Therefore R = 3x10 -3 x1x10 -3 / 1.22x 5x10 -7 = 3x10 +1 /6.1 = 30/6.1= 5 The maximum distance that can be resolved = R = 5m . Ans: b

6. 3. In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to, 2 D (b) D/ 2 (c) 2 D (d) D / 2 In young’s double slit experiment, the band width is given by β = λ D/d λ – Wavelength of light d- Slit separation D – distance between the slit and the screen β – Band width ( fringe spacing) So when d is doubled D must also be doubled ( ie . 2D) Ans: a  

7. 4. Two coherent monochromatic light beams of intensities I and 4 I are superposed. The maximum and minimum possible intensities in the resulting beam are 5I and I (b) 5I and 3I (c) 9I and I (d) 9I and 3I I max = I 1 + I 2 + 2 I 1 I 2 I min = I 1 + I 2 - 2 I 1 I 2 I max = Maximum intensity of the resultant beam I min = Minimum intensity of the resultant beam I 1 = Intensity of first beam = I I 2 = Intensity of first beam = 4I  

8. Contd … Substituting the values we get I max = I + 4I + 2 Ix4I = 5I + 2 = 5I + 2 I max I min = I + 4I - 2 Ix4I = 5I - 2 = 5I - 2 I min ie . The maximum and minimum possible intensities in the resulting beam are 9I and I Ans: c  

9. 5. When light is incident on a soap film of thickness 5×10 –5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be, 1.22 (b) 1.33 (c) 1.51 (d) 1.83 2µd = (2n + 1) λ /2 ie . µ = (2n + 1) λ /2x2d µ - Refractive index of the film d – thickness of the fil = 5×10 –5 cm = 5×10 –7 m λ - wavelength of light = 5320 Å. = 5320 10 –10 = 5.320 10 –7 m For visible region n = 2

10. Contd … Substituting the values we get µ = (2x2 + 1) 5.320x 10 –7 /2x2x 5×10 –7 m µ = 5 x 5.320x 10 –7 /4x 5 × 10 –7 = 5.32/4 = 1.33 = µ Ans: b

11. 6. First diffraction minimum due to a single slit of width 1.0×10 –5 cm is at 30 o . Then wavelength of light used is, 400 Å (b) 500 Å (c) 600 Å (d) 700 Å a sin θ = m λ ie . λ = a sin θ /m m- order of the spectrum = 1 a – Width of the slit = 1.0×10 –5 cm = 1.0×10 –7 m θ – Angle of diffraction = 30 and sin 30 = 1/2 ie . λ = 1.0×10 –7 x1/2x1 = 0.5x 10 –7 m = 500×10 –9 m = 500nm Ans: b

12. 7. A ray of light strikes a glass plate at an angle 60 o . If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is, (a) 3 (b) 3/2 (c) (3/2) (d) 2 N P Q i A O B r N 1 R   From the fig. PO – incident ray OQ-Reflected ray OR – Refracted ray NN 1 = Normal T he reflected and refracted rays are perpendicular to each other ie . ∠ QOR = 90 o .

13. Contd … By law of reflection ∠PON = i – Angle of incidence = ∠NOQ - Angle of reflection = 60 o . From the fig. ∠ NON 1 = 180 o . = ∠NOQ + ∠QOR + ∠RO N 1 ie . 180 o . = 6 o + 9 o . + r ie . r = 180 o . - 6 o - 9 o = 3 o - Angle of refraction

14. Contd … By Snell’s law µ = sin i /sin r = sin 60 o / sin 3 o = ie . µ = The refractive index of the glass µ = Ans: a  

15. 8. One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will, get shifted downwards get shifted upwards (c) will remain the same (d) data insufficient to conclude When a slit is covered with a glass plate, the path difference will increase. We know that y = n λ D/d. So the entire fringe system will move upwards . Ans: b

16. 9. Light transmitted by Nicol prism is, (a) partially polarised (b) unpolarised (c) plane polarised (d) elliptically polarised Nicol prism is an optical device used to produce plane polarised light Ans: c

17. 10. The transverse nature of light is shown in, (a) Interference (b) diffraction (c) scattering (d) polarisation Polarisation is the phenomena that proves the transverse nature light Ans: d

18. Thank You