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

1. Our eyes see two objects as separate, only in the angle subtended by them at the eye is greater than ______________

a) 30 seconds

b) 1 minute

c) 2 minute

d) 10 seconds

View Answer

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.

2. In Rayleigh’s Criterion for Resolution, two images would be just resolved when _______

a) The central maxima of one image coincide with central maxima of the other

b) The central maxima of one do not coincide with central maxima of the other

c) The central maxima of one image coincides with the first minimum of the other

d) The central maxima of one image do not coincide with the first minimum of other

View Answer

Explanation: In Rayleigh’s Criterion, the diffraction pattern formed is considered for resolution. According to him, two nearby images would be just resolved when the central maxima of one image coincide with the first minimum of the other and vice versa.

3. What is the SI unit of Resolving power?

a) m^{-1}

b) cm^{-1}

c) s^{-1}

d) no SI unit

View Answer

Explanation: Mathematically, resolving power 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.

4. The resolving power of a telescope is directly proportional to ______

a) Frequency of the light used

b) The wavelength of the light used

c) Square of the frequency of light used

d) Square of Wavelength of the light used

View Answer

Explanation: As we know, the resolving power of a telescope = D/1.22λ

Now, as we know λ = c/v, where v is the frequency of light.

Therefore, resolving power of a telescope = Dv/1.22

Hence, Resolving Power is directly proportional to the frequency of the light.

5. The resolving power of a grating is directly proportional to grating constant.

a) True

b) False

View Answer

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. There are three prisms A, B, C of base width 5 cm each with an angle of prism 30°, 36°, 45° respectively. If the light of wavelength 5000A is incident on each of them, which has the highest resolving power?

a) A

b) B

c) C

d) All of them have the same resolving power

View Answer

Explanation: The resolving power of a prism is directly proportional to the width of the base of the prism and inversely proportional to the cube of the wavelength of light. It is independent of the angle of prism. Thus, as the base width and the wavelength of light are the same in each case, all of them will have the same resolving power.

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

View Answer

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

View Answer

Explanation: Now, we know that for a grating \(\frac{\lambda}{d\lambda}\)=nN

λ

_{mean}= 5893 Å = 5.89 X 10

^{-7}m

dλ = 6 Å = 6 X 10

^{-10}m

n = 1, N =?

So, N = \(\frac{\lambda}{nXd\lambda}\) = 982

No of lines per cm = 982/2.5 = 392.

9. A microscope of objective focal length 1 cm and an eyepiece of focal length 2.0 cm has a tube length of 20 cm. What will be the magnification of the microscope?

a) 40

b) 100

c) 200

d) 250

View Answer

Explanation: We know for a microscope, magnification = (L/f

_{o}) X (D/f

_{e})

Here, f

_{o}= 1 cm and f

_{e}= 2 cm. Therefore, D = 25 cm and L = 20cm.

Hence, magnification = 20 X 25/2 = 250.

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

View Answer

Explanation: N = 16000 X 5 = 80000, n = 2

Wavelength = 6 X 10

^{-5}cm

Resolving power = nN = 2 X 80000 = 160000

Smallest wavelength difference = λ/nN

= 0.0375 Å

11. How are these two images resolved?

a) Not resolved

b) Just Resolved

c) Partially resolved

d) Well resolved

View Answer

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, the angular separation between the two central maxima is large enough for the images to be seen clearly. Thus, the images are well-resolved.

12. What is the relationship between the focal length, f, of the objective and the resolving power of the telescope, R.P?

a) f ∝ R.P

b) f ∝ 1/R.P

c) f^{2} ∝ R.P

d) no relation

View Answer

Explanation: For a telescope, the radius of the first dark ring = 1.22fλ/D

Therefore, the smaller will be the focal length of the objective, the sharper will be the diffraction pattern and hence the resolving power of the telescope would increase.

**Sanfoundry Global Education & Learning Series – Engineering Physics.**

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