Engineering Physics Questions and Answers – Dispersion, Fermat’s Principle & its Application

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Dispersion, Fermat’s Principle & its Application”.

1. The product of the Geometric Length of the path light follows through the system, and the refractive index of the medium is known as __________
a) Fermat Path
b) Optical Path
c) Ray Path
d) Optical Path Difference
View Answer

Answer: b
Explanation: The product of refractive index of the medium & geometric distance travelled by the light in the medium is known as Optical Path. A difference in optical path length between two paths is called Optical Path Difference. An Imaginary path along which energy travels associated with a point on a wave front is known as Ray Path.
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2. What will be the time taken by a ray of light travelling from one point to another, by any number of refractions & reflections follows that particular path?
a) Highest
b) Least
c) Maximum
d) Moderate
View Answer

Answer: b
Explanation: In different materials light travels at different speed. When a beam of light passes from one material to another it bends in such a way, that the path it takes from one point to another, requires the least possible time.

3. Name the person, who formed “The Principle of least time“?
a) Pierre de Fermat
b) C.V Raman
c) John Kepler
d) Albert Abraham Michelson
View Answer

Answer: a
Explanation: Fermat’s Principle or the principle of least time was named after French mathematician Pierre de Fermat. This principle states that the path taken between two points is the path that can be traversed in the least time.

4. What does first law of reflection state?
a) The incident ray, reflected ray and normal ray, at the point of incidence, are in same plane
b) The incident ray, reflected ray and normal ray, at the point of incidence, are perpendicular to each other
c) The incident ray, reflected ray and normal ray at the point of incidence are in different plane
d) mSin(i)=nSir(r)
View Answer

Answer: a
Explanation: According to the first law of reflection incident ray, reflected ray and normal ray, at the point of incidence, are in same plane. Whereas mSin(i)=nSin(r) is Snell’s Law where m & n are Refractive index of different medium.

5. If a monochromatic light goes through 2 cm of glass or 2.25 cm of water, the optical path is the same. If refractive index of glass is 1.50, then what is the refractive index of water?
a) 1.33
b) 1.23
c) 1.50
d) 1.13
View Answer

Answer: a
Explanation: The product of refractive index of the medium & geometric distance travelled by the light in the medium is known as Optical Path.
=> nA=mB (m&n are refractive index of water & glass respectively and A&B are geometric distance of glass & water)
=> 2*1.50=m*2.25
=> m=1.33.
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6. If light travels in a medium of Refractive Index 1.4 through a distance 250 m in a given time interval, then how much distance will it cover in refractive index 1.75?
a) 250 m
b) 500 m
c) 200 m
d) 312.5 m
View Answer

Answer: c
Explanation: The product of refractive index of the medium & geometric distance travelled by the light in the medium is known as Optical Path.
=> nA=mB (m&n are Refractive index of 1st & 2nd respectively and Geometric distance of 2nd & 1st)
=> n*1.75=1.4*250
=> n=200.

7. White light is passed through a prism having angle 5°. If the refractive indices for red color and blur color are 1.641 and 1.659 respectively. Find the angle between them.
a) 3.295°
b) 4°
c) 0.090°
d) 0.65°
View Answer

Answer: c
Explanation: For small angle of prism, deviation is given by;
α=(µ-1)*A
for blue color:- α1 = (1.659-1)*5° = 3.295° and
for red color:- α2 = (1.641-1)*5° = 3.205° so
therefore, Ө = α1 – α2 = 3.295°-3.205° = 0.090°.

8. A thin prism P1 having angle 4° and refractive index 1.54 is combined with another thin prism P2 of refractive index 1.72 for producing dispersion without deviation. The angle of P2 is?
a) 4°
b) 5.33°
c) 3°
d) 2.6°
View Answer

Answer: c
Explanation: Since, (µ1 – 1)A1 = (µ2 – 1)A2
(1.54 – 1)*4° = (1.72 – 1)*A2
0.54*4° = 0.72*A2
therefore, A2 = 2.16 / 0.72 = 3°.

9. A crown glass prism with refractive index 6° is to be achromatized for red and blue color light with flint glass prism. What will be the angle of flint glass prism? (µr = 1.513, µr‘ = 1.645, µ2 = 1.523 and µ2‘ = 1.665, Where µr and µ2 are refractive indices of red and blue light respectively whereas, µr‘ and µ2‘ are refractive indices of red and blue light respectively for flint glass.)
a) 1.5°
b) 2.4°
c) 3°
d) 4.5°
View Answer

Answer: c
Explanation: (µ2 – µr)*A1 = (µ2‘ – µr‘)*A2
(1.523 – 1.513)*6° = (1.665 – 1.645)*A2
0.010*6° = 0.020*A2
A2 = 3°.
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10. Find the ratio of mean deviation produced, if the ratio of dispersive power of the material of two prism are 2:3 and ratio of angular dispersions produced by them is 1:2.
a) 1:3
b) 4:3
c) 3:1
d) 3:4
View Answer

Answer: d
Explanation: let ω be the mean deviation produced,
ω = (angular dispersion) / dispersive power
ω = (1:2 ) / (2:3)
therefore, ω = 3:4.

11. Dispersive power of the material of the prism is 0.0221. If the deviation produced by it for yellow color is 38°, then the angular dispersion between red and violet color is?
a) 0.84°
b) 0.65°
c) 0.48°
d) 1.26°
View Answer

Answer: a
Explanation: Angular dispersion = deviation produced * dispersive power
angular dispersion = (38°) * (0.0221)
therefore, angular dispersion = 0.8398° Ξ 0.84°.

Sanfoundry Global Education & Learning Series – Engineering Physics.

To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn