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1.

A parallel beam of white light is incident normally on a water film` 1.0 x 10^-4` cm thick. Find the wavelength in the visible range (400 nm-700 nm) which are strongly transmitted by the film Refractive index of water = 1.33.

Answer» Correct Answer - A::B::C::D
for strong transmission
`2mud=nm`
`rarr l=2(mud)/n`
Given that `mu=1.33,
`d=1xx10^-6 cm`
`=d=1xx10^-6m`
`- lamda=(2xx1.33xx1xx10^-6)/n`
`=(2660xx10^-9)/n m`
when `n=4, `lamda_1=665nm`
`n=5`
`lamda_2=532nm`
`n=6`
`lamda_3=443 nm`
Discussion
2.

A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed ?

Answer» Correct Answer - A::C
`lamda =620nm =620xx10^-9m,
D=20cm=20xx10^-2m,
d=8cm=8xx10^-2m`
`:.R=1.22xx(620xx10^-9xx20xx10^-2)/(8xx10^-2)`
`=1891xx10^-9`
`=1.9xx10^-6m`
So, diameter `=2R=3.8xx10^-6m`
Discussion
3.

In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is `5.0 x 10^ -7 ` m and the distance of the screen from the slits is 1.0 m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimeter width on the screen ?

Answer» Correct Answer - A::B
Givent that `d=1 m=10^-3m,`
`lamda=5xx0^-7m, D=1m`
`So, Fringe width `=(Dlamda)/d=0.5mm`
a. so, distance of center of first minimum from center of cental maximum
`=0.5/2mm=0.25mm`
b. No of fringes=10/0.5=20`
Discussion
4.

A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

Answer» Correct Answer - B
We know that the first maximum (next to central maximum) occurs at `y=(lamdaD)/d`
`Given that lamda_1=480nm`
`lamda_2=600nm`
D=150cm=1.5m`
and `d=0.25mm`
`=0.25xx10^-3m`
so, `y_1=(Dlamda_1)/d`
`=((1.5)xx480xx10^-9)/(0.25xx10^-3)`
`=2.88mm`
`y_2=((1.5)xx600xx10^-9)/(0.25xx10^-3)`
so the separation between these two bright fringes is given by
`:. Separation `=y_2-y_1`
`=3.60-2.88=0.72mm.
Discussion
5.

A double slit `S_1-S-2` is illuminated by a coherent light of wavelength `lamda`. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance `D_1`, from it and a screen `sum` is placed behind the double slit at a distance `D_2`, from it . The screen `sum` receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.

Answer» Correct Answer - A::B::D
The apparent distance of the screen from the slits is
`D=2D_1+D_2`
`So fringe width= (Dlamda)/d=((2D_1+D_2)lamda)/d`
Discussion
6.

Find the thickness of a plate which will produce a change in optical path equal to half the wavelength `lamda` of the light passing through it normally. The refractive index of the plate is `mu`.

Answer» Correct Answer - A::B::D
Let t=thickness of the plate
Gien optical path difference
`=(mu-1)t=lamda/2`
`rarr t=lamda/(2(mu-1))`
Discussion
7.

Two narrow slits emitting light in phase are separated by a distance of 1-0 cm. The wavelength of the light is `5.0 x 10^-7` m. The interference pattern is observed on a screen placed at a distance of 1-0 m. (a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima ? (b) Find the separation between the sources which will give a separation of 1.0 mm between the consecutive maxima.

Answer» Correct Answer - A::B
Given that `d=1cm=10^-2m`
`=lamda=5xx10^-7m and D=1m`
a. Separation between two consecutive maxima is equal to fringe to fringe width.
so, `beta=(lamdaD)/d`
`=(5xx10^-7xx1)/10^-2m`
`=5xx10^I-5m=0.05mm`
b. When `beta=1mm=10^-3m`
`=10^-3m=(5xx10^-7xx1)/d`
`rarr d=5xx10^-4m`
`=0.50mm`.
Discussion
8.

Consider the situation of the previous problem, if the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen ?

Answer» Correct Answer - A
Given that the miror reflects 64% of energy (intensity) of light
So, `I_1/I_2=0.64=16/25`
`=r_1/r_2=4/5`
So, `I_(max)/I_(min)=((r_1+r_2)^2)/((r_1-r_2)^2)`
`=((4+5)^2)/(4-5)^2)=81:1`
Discussion
9.

A narrow slit S transmitting light of wavelength `lamda` is placed a distance d above a large plane mirror as shown in figure. The light coming directly from the slit and that coming after the reflection interference at a screen `sum` placed at a distance LD from the slit. a. What will be the intensity at a point just above the mirror. i.e., just above O? b. At what distance from O does the first maximum occur? ?

Answer» Correct Answer - A::B::D
a. Since there is a phase difference of `pi` between direct light and reflecting light the intensity just above the mirror will be zero.ltbr. B. Here 2nd= equivalent slit seperationn
D=distance between slit and screen.
We know for bright fringe
`/_x=(yxx2d)/D=nlamda`
But as thereis a phase reversal of `lamda/2`
`rarr (yxx2d)D=lamda/2=nlamda`
`rarr (yxx2d)/D=n lamda=lamda/2((n-1)/2)=lamda/2`
`rarr y=(lamdaD)/(4D)`
Discussion
10.

A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a YDSE. The paper transimits `4//9` of the light energy falling on it. a. Find the ratio of maximum intensity to the minimum intensity in interference pattern. b. How many fringes will cross through the center if an indentical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.

Answer» Correct Answer - A::B
Given that `t=0.02mm`
`=0.02xx10^-3m`
`mu_1=1.45`
`lamda=600nm=600xx10^-9m`
a. `Let I_1=` intensity of source without paper =I
Then `I_2` = intensity of source with paper
`=(1/9)I`
`rarr I_1/I_2=9/4`
`rarr r_1/r_2=3/2 [:. Ipropr^2]`
`where r_1 and r_2` are corresponding amplitudes
`So, I_(max)/I_(min)=((r_1+r_2)^2)/((r_1+r_2)^2)=((3+2)^2)/((3-2)^2)`
`=25/1=25:`
b. No of fringe that will cross the origin is given by
`n=(mu-1)t)/lamda`
`=((1.45-1)xx0.02xx10^-3)`
`=(0.45xx0.02x10^-3)/(6xx10^-7)=15`
Discussion
11.

The wavelength of sodium light in air is 589 nm. (a) Find its frquency in air. (b) Find its wavelength in water (refactive index = 1.33). (c ) find its frequency in water : (d) Find its speed in water.

Answer» Correct Answer - A::B::C::D
Given that, for sodium light,
`1 = 589nm = 589 xx 10^(-9) m`
(a) `f_a = (3xx10^8)/(589xx 10^(-9))`
`= 5.09 xx 10^(14) sec^(-1)`
(b) `mu_a = (lambda_w)/(lambda_q)`
`rArr (1)/(1.33) = (lambda_w)/(589 xx 10^(-9))`
`rArr = lambda_w = 443 nm`
(c )` f_w = f_a`
`= 5.09 xx 10^(14) sec^(-1)`
[Frequency does not change]
(d) `(mu_a)/(mu_w) = (upsilon_w)/(upsilon_a)`
`= rarr upsilon_w = (mu_a upsilon_a)/(mu_w)`
`=(3xx 10^8)/((1.33)) = 2.25 xx 10^8 m//sec.`
Discussion
12.

Find the minimum thcknessof a film which will strongly reflect the light of wavelength 589 nm. The refractive index of the material of the film is 1.25.

Answer» For strong reflection,the least optical path difference introduced by the film should be `lamda/2`. The optical path difference between the waves reflected from the two surfaces of the film is `2mud`. Thus, for strong reflection,
`2mud=lamda/2`
`or, d=lamda/(4lamda)=(589nm)/(4xx1.25)=118nm.
Discussion