Get important MCQs (with Answers) for CBSE Class 12th Physics Board Exam 2020 (Chapter 10 - Wave Optics). Students preparing for CBSE Class 12th Physics Board Exam usually ask about important MCQs (Multiple Choice Questions) & in this article, we have provided important questions (with answers), based on Chapter 10 (Wave Optics) of Class 12th Physics NCERT textbook. Here you will also get important links to access some important articles for the preparation of CBSE 12th board exams 2020.

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**Important MCQs for ****CBSE ****Class 12 Physics**** Board Exam 2020 (Chapter 10 - Wave Optics): **

**Q1.** In young’s double slit experiment, the width of the fringes obtained when light of wavelength 6000 Ȃ is 2. 0 mm. What will be the fringe width if the entire apparatus is immersed in a liquid of refractive index 1.33?

(a) 0.5 mm

(b) 1.0 mm

(c) 1.5 mm

(d) 2.0 mm

**Sol: **(c)

Let* β* be the fringe width when the entire apparatus is in air and *β ^{’}* be the fringe width when the whole apparatus is inside water.

**Q2.** A monochromatic light of wavelength 5100 Ȃ from a narrow slit is incident on a double slit. If the overall separation of 10 fringes on a screen 200 cm away is 2 cm then, the separation between the slit is

(a) 1.10 mm

(b) 0.51 mm

(c) 0.22 mm

(d) 0.11 mm

**Sol: **(b)

**Q3.** Consider interference between two waves from two sources of intensities *I* and *4I*. What is the ratio of intensities at the point where the phase difference is π?

(a) *I*

(b) 9 *I*

(c) 5* I*

(d) 9 *I*

**Sol: **(a)

The resultant intensity at a point where two waves of amplitudes *a _{1}* and

*a*meet with a phase difference of

_{2}*ϕ*is given by:

*I*=

_{R}*a*

_{1}^{2}+

*a*

_{2}^{2}+2

*a*cos

_{1}a_{2}*ϕ*.

Given, *a _{1}* =

*I*,

*a*= 4

_{2}*I*and

*ϕ*= π

Intensity of a wave is directly proportional to square of its amplitude

**Q4.** Two periodic waves of intensities *I _{1}* and

*I*pass through a region at the same time in the same direction. The sum of maximum and minimum intensities is:

_{2}(a) (*I*_{1} + *I*_{2})

(b) 2 (*I*_{1} + *I*_{2})

(c) (*I*_{1} ‒ *I*_{2})

(d) 2 (*I*_{1} ‒ *I*_{2})

**Sol: **(b)

The maximum intensity of two waves having intensity* I _{1}* and

*I*is given by:

_{2} The minimum intensity of two waves having intensity* I _{1}* and

*I*is given by:

_{2}Therefore, the sum of maximum and minimum intensities is: *I*_{max} + *I*_{min} = 2 (*I*_{1} + *I*_{2}).

**Q5.** The polarizing angle for a medium is 60^{o}. The critical angle for this medium is

(a) sin^{-1} (0.321)

(b) sin^{-1} (0.577)

(c) sin^{-1} (0.732)

(d) sin^{-1} (1)

**Sol: **(b)

From Brewster’s law, the refractive of the medium is given by, *μ* = tan *i _{p}* = tan 60

^{o}= 1.732.

Let C be the critical angle for the medium then,

*μ* = 1/ sin C

⇒ sin C = 1/* μ* = 1/1.732 = 0.577

⇒ C = sin^{-1} (0.577).

**Q6.** Two crossed Polaroids A and B are placed in the path of a light beam. In between these, a third Polaroid C is placed whose polarization axis makes an angle θ with the polarization axis of the Polaroid A. If *I _{o}* be the intensity of light coming out if the Polaroid A then, the intensity of light coming out of the Polaroid B is

**Sol: **(d)

Between the Polaroids A and B, a third Polaroid C is placed whose polarization-axis makes an angle θ with the polarization axis of the Polaroid A.

Let the intensity of plane polarised light emerging from A be *I _{o}* which falls on Polaroid

*C*, then by law of Malus , the intensity of light emerging from polaroid

*C*is,

*I*=

_{C}*I*cos

_{o}^{2}

*θ*.

Now this light falls on Polaroid B, whose polarizing axis makes an angle of (90^{o} ‒ *θ*) with the axis of Polaroid C.

Applying Malus law again we have,

**Q7.** A monochromatic light of wavelength 6000 Ȃ is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?

(a) 6.4 × 10^{‒7} radian

(b) 4.1 × 10^{‒7} radian

(c) 2.9 × 10^{‒7} radian

(d) 1.1 × 10^{‒7} radian

**Sol: **(c)

Given, diameter of objective of telescope = D = 100 inch = 100 × 2.54 = 254 cm

Wavelength of light = λ = 6000 Ȃ = 6000 × 10^{‒8} cm = 6 × 10^{‒5} cm

**Q8. **Light incident on a transparent medium is found to be completely plane-polarised after reflection. If the refractive index of glass plate is 1.57 then, the refracting angle of glass is:

(Given, tan^{-1}1.57 = 57.5°)

(a) 15.8^{0}

(b) 32.5^{0}

(c) 41.6^{0}

(d) None of these

**Sol: **(b)

As the reflected light is plane polarised, therefore the incident angle is polarizing angle.

Given, refractive index of the glass = *μ* = 1.57, angle of refraction = *r* =?

If, *i _{p}*is the polarizing angle then, by Brewster’s law

**Q9.** Two coherent monochromatic light beams of intensities *I* and 4*I* are superimposed. The maximum and minimum possible intensities in the resulting beam are

(a) 9 *I* and *I*

(b) *I *and 9 *I*

(c) 9 *I*and 9 *I*

(d) 3 *I* and 6 *I*

**Sol: **(a)

Let intensities be *I _{1}*,

*I*and their amplitude is

_{2}*a*

_{1}and

*a*

_{2}respectively.

**Q10.** In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5×10^{-2} m towards the slits, the change in the fringe width is 3×10^{-5} m. If the distance between the slits id 10^{-3} m, The wavelength of the light used is:

(a) 5000 Å

(b) 3000 Å

(c) 4500 Å

(d) 6000 Å

**Sol: **(d)

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