Diffraction MCQ Level - 1

10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Diffraction MCQ Level - 1

This mock test of Diffraction MCQ Level - 1 for IIT JAM helps you for every IIT JAM entrance exam. This contains 10 Multiple Choice Questions for IIT JAM Diffraction MCQ Level - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Diffraction MCQ Level - 1 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Diffraction MCQ Level - 1 exercise for a better result in the exam. You can find other Diffraction MCQ Level - 1 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.

What changes are observed in a diffraction pattern if the whole apparatus is immersed in water? 


Explanation: As the whole apparatus is now immersed in water, the wavelength of the light will change. λ’= λ / μ
Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease. Width of central maxima = 2λ / a
Therefore, as the wavelength decreases, the width of the central maxima decreases.


How does the width W of the central maximum formed from diffraction through a circular aperture (pupil) change with aperture size (b) for a fixed distance away from the aperture?


Width of central maxima is inversely proportional to slit width or aperture size .

The correct answer is: W decreases as b increases


A grating is able to resolve two very close spectral lines of wavelength 5890 and 5896  in its first order diffraction. The resolving power of the grating is :


The two lines will be seen distinctly if they are resolved by grating.
Resolving power

∴ The correct option is close to 1000.
The correct answer is: close to 1000


The diffraction pattern due to circular aperture consists of :


The correct answer is: a bright central disc surrounded by alternate dark and bright concentric rings


The distance between the maxima of two wavelengths seen on a screen, after passing through a diffraction grating :


dsin(theta) = m (lambda)

If distance between the maximas increases it means sin(theta) will increase so to normalize that d should decrease. It means that ‘d’ should decrease which in turn means that the number of slits per cm will increase.


According to Rayleigh’s criterion of resolution, the two spectral lines of equal intensity are just resolved when the central maximum of the diffraction pattern due to one falls :


The correct answer is: on the first minimum of the diffraction pattern of other


Dispersive power of a grating can be defined as :


The correct answer is: Increase of angle of diffraction w.r.t. change in wavelength


The resolving power of a grating :


The correct answer is: increases as the total number of lines on the grating increases


Which one of the following plane transmission grating of width W, and number of lines per cm, L, will have the maximum resolving power in the first order?


Resolving power = nN, where n is order of diffraction and N is the total number of ruling on grating.
N = WL
∴ R.P  = nWL

where n = 1
R.P= 1 x 5000 = 5000
R.P= 1.5 x 4000 = 6000
R.P= 2 x 2400 = 4800
R.P= 3 x 1500 = 4500
The correct answer is: W = 1.5 cm, L = 4000


If N is the total number of rulings on the grating, m is the order of spectrum and A is the wavelength of light used, then resolving power of grating is given by :


The correct answer is: mN

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