What wavelength of light would you be able to resolve at the furthest distance based on diffraction?
∴ Blue light (A = 450 nm)
The correct answer is: Blue λ = 450 nm
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.
∴ 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.P1 = 1 x 5000 = 5000
R.P2 = 1.5 x 4000 = 6000
R.P3 = 2 x 2400 = 4800
R.P4 = 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