A beam of light consisting of two wavelengths 6500 A and 5200 A is u

Question 1. In an interference pattern

16^{t h}

order maximum corresponds to 600nm . What order will be visible if 480nm is used in the same position:

5
10
15
20

Answer: Option D
: D

n = 16, y = \frac{n λ D}{d} = \frac{n^{'} λ^{'} D}{d}

n λ = n^{'} λ^{'}

\Rightarrow 16 \times 600 = n^{'} \times 480

\Rightarrow n^{'} = 20

Question 2. Two slits spaced 0.25 mm apart are placed at 0.75 m from a screen and illuminated by coherent light with a wavelength of 650 nm. The intensity at the centre of the central maximum

(θ = 0^{\circ})

is

I_{0}

. The distance on the screen from the centre of central maximum to the point where the intensity has fallen to

\frac{I_{0}}{2}

is nearly

0.1 mm
25 mm
0.4 mm
0.5 mm

Answer: Option D
: D

I = I_{0} c o s^{2} \frac{(π d s i n θ)}{λ} = \frac{I_{0}}{2}

c o s (\frac{π d s i n θ}{λ}) = \frac{1}{\sqrt{2}}

we get,

x = \frac{λ D}{4 d} = \frac{650 \times 10^{- 9} \times 0.75}{4 \times 0.25 \times 10^{- 3}} = 487.5 \times 10^{- 6} m

= 0.4875 m m = 0.5 m m

Question 3. In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is

λ

is K unit;

λ

being the wavelength of light used. The intensity at a point where the path difference is

\frac{λ}{4}

will be:

K4
K2
K
zero

Answer: Option B
: B

K_{1} = I + I + 2 I c o s δ

or

K_{1} = 2 I, K_{1} = \frac{K}{2}

For path difference

\frac{λ}{4}

, phase difference,

δ = \frac{2 π}{λ} . \frac{λ}{4} = \frac{π}{2}

Question 4. Monochromatic green light of wavelength 5×10-7 m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is

0.25 mm
0.1 mm
1.0 mm
0.01 mm

Answer: Option C
: C

Question 5. In Young's double slit experiment, if one of the slit is closed fully, what will be visible on the screen?

A slit-shaped bright spot
A YDSE interference pattern
A diffraction pattern
Nothing will be visible

Answer: Option C
: C

If one of theslits is closed then YDSE interference fringes are not formed on the screen, so you might be tempted to say 'A slit-shaped bright spot' will beall that you can see on the screen, but no, think about what you just learnt, a fringe pattern is observed due to diffraction from the single slit that is open.

Question 6. In a Young's double - slit experience, the slits are 2mm apart and are illuminated with a mixture of two wavelengths

λ_{0} = 750 n m

and

λ = 900 n m

. The minimum distance from the common central bright fringe on a screen 2 m away from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other, is

1.5mm
3 mm
4.5 mm
6 mm

Answer: Option C
: C

y_{0} = n [\frac{D λ}{d}]

and

y_{n}^{'} = n^{'} (\frac{D λ^{'}}{d})

Equating

y_{n}

and

y_{n}^{'}

,we get

\frac{n}{n^{'}} = \frac{λ^{'}}{λ} = \frac{900}{750} = \frac{6}{5}

Hence, the first position at which overlapping occurs is

y_{6} = y_{6} = \frac{6 (2) (750 \times 10^{- 9})}{2 \times 10^{- 3}} = 4.5 m m

	A.	1.1 mm
	B.	2.6 mm
	C.	2.8 nm
	D.	3.12 mm

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