# Engineering Physics Questions and Answers – Resolving Power of Grating

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Resolving Power of Grating”.

1. Which of the following is the correct expression for the resolving power of a grating?
a) (nN + 1)/λ
b) nN/λ
c) nN/λ + 1
d) nN

Explanation: The expression for the resolving power of a grating is given by: $$\frac{\lambda}{\partial\lambda}$$ = nN
Hence, it is proportional to the number of slots on the grating and the order of the spectrum. It is independent of the grating constant.

2. What is the SI unit of the Resolving power of a plane transmission grating?
a) m-1
b) cm-1
c) s-1
d) No SI unit

Explanation: Mathematically, the resolving power of a plane transmission grating can be defined as the ratio of the mean wavelength of a pair of spectral lines and the wavelength difference between them. As both the quantities have the same unit, resolving power has no unit.

3. The resolving power of a grating is inversely proportional to the wavelength of the incident light.
a) True
b) False

Explanation: As we know, the resolving power of a grating = nN
Now, n = (e + d) sinΘ / λ, where e + d is the width of the ruled surface of the grating.
Thus, Resolving power = N (e + d) sinΘ / λ.
Hence, the resolving power is inversely proportional to the wavelength of incident light.
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4. Our eyes see two objects as separate, only if the angle subtended by them at the eye is greater than _________
a) 30 seconds
b) 1 minute
c) 2 minute
d) 10 seconds

Explanation: Human eyes can see two objects as separate only when the angle subtended by them on the eye is greater than 1 minute. This is the minimum angle of resolution of a normal human eye. This process of observing two objects as separate is called resolution.

5. The resolving power of a grating is directly proportional to grating constant.
a) True
b) False

Explanation: For a grating, the resolving power = nN, where n is the order of the spectrum and N is the total number of slits on the grating. It is independent of the grating constant.

6. If a light is incident on a grating with 5000 lines/cm, then the angular separation of the two lines (5000 Å and 5006 Å) in first order spectrum is ______
a) 0.01°
b) 0.02°
c) 0.03°
d) 0.04°

Explanation: Here, N = 5000. Therefore, a + b = 1/5000 cm
Now we know, (a + b) sinΘ = nλ
For λ = 5000 Å, sinΘ1 = 5 X 10-5 X 5000
Θ1 = 14.47°
For λ = 5006 Å, sinΘ2 = 5.006 X 10-5 X 5000
Θ2 = 14.49°
Angular separation = Θ2 – Θ1 = 0.02°.

7. Light is incident normally on a grating of width 5 X 10-3 m with 2500 lines. What is the resolving power of the grating in the second order spectrum?
a) 2500
b) 5000
c) 1250
d) 500

Explanation: The resolving power of a grating = nN, where n is the order of the spectrum and N is the number of slits on the grating.
Thus, Resolving power = 2 X 2500
= 5000.

8. What is the minimum number of lines per cm in a 2.5 cm wide grating spectrum which will just resolve two sodium lines (5890 Å and 5896 Å) in the first order spectrum?
a) 98
b) 196
c) 392
d) 694

Explanation: Now, we know that for a grating $$\frac{\lambda}{d\lambda}$$=nN
λmean = 5893 Å = 5.89 X 10-7 m
d λ = 6 Å = 6 X 10-10m
n = 1, N = ?
So , N = $$\frac{\lambda}{n Xd\lambda}$$ = 982
No of lines per cm = 982/2.5 = 392.

9. The following pattern was observed by a plane diffraction grating. How are the two images resolved?

a) Not resolved
b) Just Resolved
c) Partially resolved
d) Well resolved

Explanation: As stated in Rayleigh’s criteria, when the central maxima of one image coincide with the first minimum of the other image, the two images are said to be just resolved. In the fig, this condition is satisfied. Hence, the two lines are just resolved.

10. A grating has 16000 per inch over a length of 5 inches. What will be the smallest wavelength difference for a light of wavelength 6000 Å?
a) 0.01 Å
b) 0.02 Å
c) 0.03 Å
d) 0.04 Å

Explanation: N = 16000 X 5 = 80000, n = 2
Wavelength = 6 X 10-5cm
Resolving power = nN = 2 X 80000 = 160000
Smallest wavelength difference = λ/nN
= 0.0375 Å.

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