InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A single slit of width 0.1mm is illuminated by parallel light of wavelength 6000Å, and diffraction bands are observed on a screen 40cm from the slit. The distance of third dark band from the central bright band is:A. 7.2 mmB. 3.6 mmC. 2.4 mmD. 0.6 mm |
|
Answer» Correct Answer - A Width `=(3xx6xx10^(-7)xx0.4)/(10^(-4)) =7.2 mm` |
|
| 102. |
Two slits separated by 0.33mm.A beam of `500nm` light strikes the slits producing an interference pattern.Tegh number of maxiam observed in the angular angle range`-30^(@)ltthetalt30^(@)` A. `300`B. `150`C. `599`D. `149` |
|
Answer» Correct Answer - C `d sintheta-nlambda` `n=(3xx10^(-4)xxsin30^(@))/(5xx10^(-7)) rArr n=300` since-`30^(@)lttheta lt 30^(@)` total number of maximum=`(600-2+1)=599` |
|
| 103. |
The intensity on the screen at a certain point in a double-slit interference patteren is 25.0% of the maximum value. (a) What minimum phase difference (in radians) between the sources produces this result? (b) Express this phase difference as a path difference for 486.1 nm light. |
|
Answer» (a) `(I)/(I_(max)) = cos^(2) ((phi)/(2))` Therefore, `phi = 2 cos^(-1) sqrt((I)/(I_(max))) = 2 cos^(-1) sqrt 0.25` `2 cos^(-1) ((1)/(2)) = 2 xx (pi)/(3) = (2pi)/(3) rad` (b) `delta = (lambda phi)/(2 pi) = ((486 nm)((2)/(3) pi "rad"))/(2 pi) = 162` |
|
| 104. |
Two slits are separated by 0.320 mm. A beam of 500 nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range `- 30.0^(@) lt theta lt 30.0^(@)`. |
|
Answer» At `30.0^(@), d sin theta = m lambda` `(3.20 xx 10^(-4) m ) sin 30.0^(@) = m (500 xx 10^(-9) m)` So, `m = 320` There are 320 maxima ot the right, 320 to the left, and one for `m = 0` straight ahead. Therefore, there are 641 maxima. |
|
| 105. |
With two slits spaced 0.2 mm apart and a screen at a distance of 1 m, the third bright fringe is found to be at 7.5 mm from the central fringe. The wavelength of light used isA. `400 nm`B. `500 nm`C. `550 nm`D. `600 nm` |
|
Answer» Correct Answer - B `y = (n lambda D)/(d)` |
|
| 106. |
A soap bubble `(n = 1.33)` is floating in air. If the thickness of the bubble wall is 115 nm, what is the wavelength of the light that is most strongly reflected? |
|
Answer» Light reflecting from the first suffer phase reversal. Light reflecting from the second surface does not, but passes twice through the thickness t of the film. So for consturctive interference, we require `(lambda_(n))/(2) + 2 t = lambda n` where `lambda_(n) = (lambda)/(n)` is the wavelength in the material. Then, `2 t = (lambda_(n))/(2) = (lambda)/(2n)` `:. lambda = 4 nt = 4 (1.33) (115nm) = 612 nm` |
|
| 107. |
What speed should a galaxy move with respect to us so that the sodium line at `589.0 nm` is observed at `589.6 nm` ? |
|
Answer» `(Delta lambda)/(lambda) = (V)/(C )` , `V= +c((0.6)/(589.0))=3xx10^(8) ((0.6)/(589.0))= +3.06xx10^(5) ms^(-1)` Therefore. the galaxy is moving away forms us with speed `306 km//s`. |
|
| 108. |
What speed should a galaxy move with respect to use so that the sodium line at 589.0 nm is oberved at 589.6 nm ? |
|
Answer» Since, `v lambda=c, (Deltav)/v=(Deltalambda)/(lambda)` (for small changes in v and `lambda)`. For `Deltalambda =589.6-589.0=+0.6nm` we get [using Equation `=(Deltav)/(v)=(upsilon_("radial"))/(c)]` `(Deltav)/(v)=-(Deltalambda)/(lambda)=-(upsilon_("radical"))/(c)` or `upsilon_("radical")cong +c ((0.6)/(589.0))=+3.06xx10^(5)m s^(-1)=306km//s` Therefore, the galaxy is moving away from us. |
|
| 109. |
A galaxy moving with speed 300km/s shows blue shift. At what wavelength sodium line at 589.0 nm will be observed ? |
|
Answer» Here, `v_("radial")=-300 km//s=-3xx10^(5)m//s` Negative sign comes because we observe blue shift i.e., galaxy is approaching us. Now, `-(Deltalambda)/(lambda)=-(v_("radial"))/c` `rArr=-(Deltalambda)/(589.0 nm)=(-(-3xx10^(5)))/((3xx10^(8)))` `rArr Delta lambda=-(589.0)xx10^(-3) nm` =-0.589 Hence, the observed wavelength will be (589.0-0.589)nm =588.411 nm |
|
| 110. |
In a Laser beam the photons emitted areA. same wavelengthB. coherentC. of same velocityD. All the above |
| Answer» Correct Answer - D | |
| 111. |
Assertion (A) : The flim which appears bright in reflected system will appear dark in the transmitted system and vice-versa. Reason (R ) : The condittions for film to appear bright of dark in the reflected light are just revese to those in the transmitted lightA. A is true and R is true and R is the correct explanation of A.B. A and R are true but R is not correct explanation of AC. A is true, R is falseD. A is false, R is true. |
|
Answer» Correct Answer - A For reflected system of the film, the condition for maxima is `2mu t cos r = (2n-1)lambda//2` . While the maxima for transmitted system of film is `2 mut cos r = n lambda` and is reverse for minima. A & R are true and R is the correct explanation of A |
|
| 112. |
Assertion (A) : The flim which appears bright in reflected system will appear dark in the transmitted system and vice-versa. Reason (R ) : The conditions for film to appear bright of dark in the reflected light are just revese to those in the transmitted lightA. (a) If both assertion and reason are true and reason is the correct explanation of assertion.B. (b) If both assertion and reason are true but reason is not the correct explanation of assertion.C. (c) If assertion is true but reason is false.D. (d) If assertion and reason both are false. |
|
Answer» Correct Answer - A For reflected system of the film, the maxima or constructive interference is `2mutcosr=((2n-1)lambda)/(2)` while the maxima for transmitted system of film is given by equation `2mutcosr=nlambda` where t is thickness of the film and r is angle of reflection. From these two equations we can see that condition for maxima in rejected system and transmitted system are just opposite. |
|
| 113. |
A Soap bubble has a thickness of 90 nm and its refractive index is `mu" = 1.4"`. What colour does the bubble appear to be at a point on its surface closest to an observer when it is illuminated by white light? |
|
Answer» Correct Answer - green |
|
| 114. |
With what speed should a galaxy move with respect to us to that the sodium line at `589.0 nm` is observed at `589.6 nm` ?A. 206 km `s^(-1)`B. 306 km `s^(-1)`C. 103 km `s^(-1)`D. 51 km `s^(-1)` |
|
Answer» Correct Answer - B The relation between v, c and `lambda` are ,`v lambda=c` For small change in v and `lambda`, `(triangle v)/(v)=(-triangle lambda)/(lambda)` As `triangle lambda =589.6-589.0=+0.6 nm` Therefore, using doppler shift `(triangle v)/(v)=(-triangle lambda)/(lambda)=(-v_("radial"))/(c)` or `-v_("radial")~=+c((0.6)/(589.0))=+3.06xx10^(5)ms^(-1)` ` " " =306kms^(- 1)` Therefore, the galaxy is moving away from us. |
|
| 115. |
A laser beam is used for locating distant objects becauseA. it is monochromaticB. it is not chromaticC. it is not observedD. it has small angular spread |
|
Answer» Correct Answer - D A laser beam is used for locating distant objects because it has small abular spread. |
|
| 116. |
A plane wave of monochromatic light falls normally on a uniform thin or oil which covers a glass plate. The wavelength of source can be varied continuously. Complete destructive is observed for `lambda = 5000 Å` and `lambda = 1000 Å` and for no other wavelength in between. If `mu` of oil is 1.3 and that of glass is 1.5, the thickness of the film will beA. `6.738 xx 10(-5) cm`B. `5.7 xx 10^(-5) cm`C. `4 xx 10^(-5) cm`D. `2.8 xx 10^(-5) cm` |
|
Answer» Correct Answer - a In this case, both the rays suffer a phase change of `180^(@)` and the conditions for destructive interference is `2 n d = (m + (1)/(2)) lambda_(1)` `2 n d = (m + (3)/(2)) lambda_(2)` `:. (m + (1)/(2))/(m + (3)/(2)) = (lambda_(2))/(lambda_(1)) = (5000)/(7000) = (5)/(7)` and `d = ((m + (1)/(2))lambda_(1))/(2n) = (2.5 xx 7000)/(2 xx 1.3)` `= 6738 Å = 6.738 xx 10^(-5)cm` |
|
| 117. |
A soap flim of thickness `0.3 mu m` appears dark when seen by the refracted light of wavelength `580 nm`. What is the index of refraction of the soap sloution, if it is known to be between `1.3` and `1.5`? |
|
Answer» Correct Answer - 1.45 The path difference is `2mu t`. Now for destructive interface it can be `2mut=(lambda)/(2)` or `(3lambda)/(2)` or `(5lambda)/(2)` and so on……… `mu=(lambda)/(4 t),(3lambda)/(4 t),(5lambda)/(4 t)……..` `=(580xx10^(-9))/(4xx0.3xx10^(-6)) , (3xx580xx10^(-9))/(4xx0.3xx10^(-6)) .......` `0.4833, 3xx0.4833......` so only `mu=3xx0.4833=1.45` is the answer, `{1.3 lt mu lt 1.5}` |
|
| 118. |
The wavelength of spectral line coming from a distant star shifts from 600 nm to 600.1 nm. The velocity of the star relative to earth isA. 50 km `s^(-1)`B. 100 km `s^(-1)`C. 25 km `s^(-1)`D. 200 km `s^(-1)` |
|
Answer» Correct Answer - A Here, `lambda = 600 nm` `triangle lambda=600.1 nm - 600nm=0.1 nm` As `(v_(s))/(c)= (triangle lambda)/(lambda) , :. (triangle lambda)/(lambda)c=(0.1nm)/(600nm)xx3xx10^(8)ms^(-1)` ` :.v_(s)=50xx10^(3)ms^(-1)=50kms^(-1)` |
|
| 119. |
A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?A. (a) `4800Å`B. (b) `5800Å`C. (c) `6000Å`D. (d) `6800Å` |
|
Answer» Correct Answer - A Condition for observing bright fringe is: `2^(nd)=(m+1/2)lambda` `:. lambda=(2nd)/(m+1/2)=(2xx1.5xx4xx10^-5)/(m+1/2)=(12xx10^-5)/(m+1/2)` The integer m that gives the wavelength in the visible region (`4000Å` to `Å`) is `m=2`. In that case, `lambda=(12xx10^-5)/(2+1/2)=4.8xx10^-5=4800Å`. |
|
| 120. |
A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?A. `4800 Å`B. `5800 Å`C. `6000 Å`D. `6800 Å` |
|
Answer» Correct Answer - a Condition for observing bright fringe is `2 nd = (m + (1)/(2)) lambda` `:. lambda = (2 nd)/((m + (1)/(2))) = (2 xx 1.5 xx 4 xx 10^(-5))/(m + (1)/(2))` `= (12 xx 10^(-5))/(m + (1)/(2))` The integer m that gives the wavelength in the visible region (`4000 Å` to `7000 Å`) is `m = 2`. In that case, `lambda = (12 xx 10^(-5))/(2 (1)/(2)) = 4.8 xx 10^(-5) = 4800 Å` |
|
| 121. |
A thin film having refractive index` mu" = 1.5"` has air on both sides. It is illuminated by white light falling normally on it. Analysis of the reflected light shows that the wavelengths 450 nm and 540 nm are the only missing wavelengths in the visible portion of the spectrum. Assume that visible range is 400 nm to 780 nm. (a) Find thickness of the film. (b) Which wavelengths are brightest in the interference pattern of the reflected light? |
|
Answer» Correct Answer - (a)900 nm (b)415nm ,491nm,600nm,771nm |
|
| 122. |
A star is going away from the earth. An observer on the earth will see the wavelength of light coming from the starA. decreasedB. increasedC. neither decreased nor increasedD. decreased or increased depending upon the velocity of the start |
| Answer» Correct Answer - B | |
| 123. |
If the shift of wavelength of light emitted by a star is towards viloet, then this shows that star isA. stationaryB. moving towards earthC. moving away from earthD. Information is incomplete |
| Answer» Correct Answer - B | |
| 124. |
A thin film of thickness `4xx10^(-5)` cm and `mu`=1.5 is iluminated by white light incident normal to its surface. What wavelength in the visible range be intensified in the reflected beam ? |
|
Answer» Correct Answer - 4800 Å |
|
| 125. |
Consider a film of thickness L as shown in four different cases belew. Notice the observation of film with perpendicularly falling light. Mark the correct statememt(S). A. For (1) and (2), the reflection at film interfaces causes zero phase difference for two reflected rays.B. For (2) and (3), the reflection at film interfaces causes a phase difference of `pi` for two reflected rays.C. For (1), the film will appear dark, if it is observed through reflected rays from film interfaces.D. For (3), the film will appear dark, if it is observed through reflected rays from film interfaces. |
|
Answer» Correct Answer - b.,d It is better to make a chart for all the four cases which show the phase difference due to reflection at top and bottom surfaces of film. `{:(,underset("surfaces")("Top"),underset("surfaces")("Bottom"),"Due 2L","Total"),("Case I",0,0,6 pi,6 pi),("Case II",0,pi,6 pi,7 pi),("Case III",0,pi,6 pi,7 pi),("Case IV",pi,pi,6 pi,8 pi):}`. |
|
| 126. |
Two point sources are placed on a straight line separated by a distance `d = 3 lambda`. Both the sources are placed at a distance L from a wall which is perpendicular to the straight line. Both the sources are sending waves of equal intensity. Find: a. Locus of the points on the wall having equal intensity. b. Maximum and minimun path difference observed on the wall. |
|
Answer» As shown in the figure, path difference between the waves is same in the circular path on the wall. Hence, locus of the points of same intensity is a circle. Maximum path difference will be at `O = 3 lambda`. Minimum path difference will be at a large distance from O and it will be zero. `(BMS_V04_C02_S01_008_S01.png" width="80%"> |
|
| 127. |
Red shift is an illustration ofA. low temperature emissionB. high frequency absorptionC. Doppler effectD. unknown phenomenon |
| Answer» Correct Answer - C | |
| 128. |
White light, with a uniform intensity across the visible wavelength range 430 - 690 nm, is perpendicularly incident on a wate film, of index of refraction `mu = 1.33` and thickness `d = 320`nm, that is suspended in air. At what wavelength `lambda` is the light reflected by the film brightest to an observer? |
|
Answer» In this situtaion, Eq. (i) gives the interference maxima. Solving for `lambda` and inserting the given data, we obtain `lambda = (2 mu d)/(n + 1//1) = ((2)(1.33)(320 km))/(n + 1//2) = (851 nm)/(n + 1//2)` For `m = 0`, this give us `lambda = 1700 nm` , which is in the infrared region. For `n = 1`, we find `lambda = 567 nm`, which is yellow-green light, near the middle of the visible spectrum. For `n = 2, lambda = 340 nm`, which is in the ultraviolet region. So, the wavelength at which the light seen by the observer is brightest is `lambda = 567 nm`. |
|
| 129. |
A ray of light is incident on a thin film, As shown in figure, M and N two reflected rays while P and Q are two transmitted rays. Rays N and Q undergo a phase change of `pi`. Correct ordering of the refracting indices is A. `n_(2) gt n_(3) gt n_(1)`B. `n_(3) gt n_(2) gt n_(1)`C. `n_(3) gt n_() gt n_(2)`D. none of these, the specified changes cannot occur |
|
Answer» Correct Answer - b Ray N undergoes reflection at surface II with phase change of `pi`. `implies n_(3) gt n_(2) ` Ray Q undergoes a phase change of `pi` at surface II, but there is no phase change when it is reflected from surface I. `implies n_(1) lt n_(2) ` |
|
| 130. |
A ray of light is incident on a thin film, As shown in figure, M and N two reflected rays while P and Q are two transmitted rays. Rays N and Q undergo a phase change of `pi`. Correct ordering of the refracting indices is A. `n_(2)gtn_(3)gtn_(1)`B. `n_(3)gtn_(2)gtn_(1)`C. `n_(3)gtn_(1)gtn_(2)`D. none of these the specified changes can not occur |
|
Answer» Correct Answer - B There is a phase change of x when the ray enters from rarer medium into denser medium, the boundary is called rigid boundary and also no change in phase when it enters into rarer medium. |
|
| 131. |
(a) In Illustration 2.20, how many dark rings will be observed on the wall? (b) What is the Path difference at point P? |
|
Answer» (a) Path difference corresponding to minima in our given range is `lambda//2, 3 lambda//2, 5 lambda//2`. Hence, three dark rings will be observed on the wall. (b) Path difference between the waves reaching at P, from Fig. 2.20 `Delta x = S_(2) A` `S_(2) P - S_(1) P = Delta x or d cos theta = Delta x` `cos theta = (Delta x)/(d)` But `tan theta = (y)/(D)` `implies cos theta = (sqrt(D ^(2) - Y^(2)))/(D)` `(D sqrt(1 - (y^(2))/(D^(2))))/(D)` `= (1 - (y^(2))/(D^(2)))^(1//2) = (1 - (Y^(2))/(2D^(2)))` From (i) `Delta x = d cos theta = d (1-(y^(2))/(2D^(2)))` From (i) and (ii), we can find the radius of any dark of bright ring. |
|
| 132. |
In a biprism experiment , the two virtual images of the slit is `1.2` mm and the wavelength of light used in 4000 Å . If the distance of the 3rd bright band from the central bright band is 1 mm and the distance between the biprism and the focal plane of the eyepiece is `0.9` m , find the distance between the slit and the biprism . |
|
Answer» Correct Answer - 10 cm |
|
| 133. |
A parallel beam of monochromatic light is incident normally on a plane transmision grating having 5000 lines per cm and the second order spectral ,ine is found to be diffracted through `30^(@)`. Calculate the wavelength of light used. |
|
Answer» Correct Answer - 5000 Å |
|
| 134. |
The displacements of two interfering light waves are `y_(1) = 2sin omega t` and `y_(2) = 5sin (omega t+(pi)/(3))` the resultant amptitude isA. `39cm`B. `sqrt(39)cm`C. `7cm`D. `sqrt(29)cm` |
|
Answer» Correct Answer - B `A = sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos theta)` |
|
| 135. |
The displacements of two intering lightwaves are `y_(1) = 4 sin omega t` and `y_(2) = 3 cos(omega t)`. The amplitude of the resultant wave is (`y_(1)` and `y_(2)` are in CGS system)A. 5 cmB. 7 cmC. 1 cmD. zero |
|
Answer» Correct Answer - A `A = sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos theta)` |
|
| 136. |
Answer the questions : In what way is diffraction from each slit related to the interference pattern in a doutler slit experiment ? |
| Answer» The intensity of interference fringes in a double slit arrangement is modulated by diffraction pattern of each slit. | |
| 137. |
A flake of glass (refractive index 1.5) is placed over one of the opening of a double-slit apparatus. The interference pattern displaced itself through seven successive maxima toward the side where the flake is placed. If wavelength of the light is `lambda = 600 nm`, then the thickness of the flake isA. 2100 nmB. 4200 nmC. 8400 nmD. none of above |
|
Answer» Correct Answer - c (Shift) `= 7` (Position of central maxima) `implies (mu - 1)t (D)/(d) = 7 ((lambda D)/(d))` `:.t = (7 lambda)/(mu - 1)` ` implies t = (7(600))/(0.5)` `implies t = 8400 nm` |
|
| 138. |
In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference patternA. the intensities of both the maxima and the minima increasesB. the intensity of the maximum increase and minima has zero intensityC. the intensity of the maxima decreases and that of minima increasesD. the intensity of the maxima decreases and the minima has zero intensity |
|
Answer» Correct Answer - a When slits of equal width are taken, then intensity at maxima is `4 a^(2)` and at minima it is zero `(I prop w)`. When one slit is doubled, then intensity at maxima will increase whereas intensity at minima will no be equl to zero and will be finite. |
|
| 139. |
A certain region of a soap bubble reflects red light of vacuum wavelength `lambda = 650nm`. What is the minimum thickness that this region of the soap bubble could be have? Take the index of reflection of the soap film to be 1.41.A. `1.2 xx 10^(-7) m`B. `650 xx 10^(-9) m`C. `120 xx 10^(7) m`D. `650 xx 10^(5) m` |
|
Answer» Correct Answer - a There is air on both sides of the soap film. Therefore, the reflections of the light produces a net `108^(@)` phase shift. The condition for bright fringe is `2 t = (m + (1)/(2)) lambda_("film")` `t = ((m + (1)/(2)) lambda_("film"))/(2) = ((m + (1)/(2)) lambda)/(2n)` `= ((1)/(2) (650 xx 10^(-9)))/(2(1.41)) = 1.2 xx 10^(-7) m` |
|
| 140. |
Statement-I : In YDSE, if intensity of each source is `I_(0)` then minimum and maximum intensity is zero and `4I_(0)` respectively. Statement-II : In YDSE, energy conservation is not followed.A. Statement-I is true and Statement-II is true and Statement-II is the correct explanation of Statement-I.B. Statement-I asnd Statement-II are true but Statement-II is not the correct explanation of Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
|
Answer» Correct Answer - C In YDSE, law of converstion of energy is obeyed. |
|
| 141. |
In YDSE, when a glass plate of refractive index 1.5 and thickness t is placed in the path of one of the intefering beams (wavelength`lambda`), intensity at the position where central maximum occurred previously remains unchanged. The minimum thickness of the glass plate isA. `2 lambda`B. `(2 // 3) lambda`C. `lambda//3`D. `lambda` |
|
Answer» Correct Answer - a If after placing the plate, intensity at the position of central maxima position remains unchanged, then it means first maxima takes position of central maxima, In case of minimum thickness of plate, 2 path difference created by the plates should be equal to `lambda`. i.e., `t (mu - 1) = lambda` `t ((3)/(2) - 1) = lambda implies t =2 lambda` |
|
| 142. |
Two sources with intensity `9I_(0)` and `4I_(0)` interfere in a medium. Then find the ratio of maximum to the minimum intensity in the interference pattern. |
|
Answer» `I_"max"=(sqrt(9I_0)+ sqrt(4I_0))^2=25I_0` `I_"min"=(sqrt(9I_0)-sqrt(4I_0))^2=I_0` `I_"max"/I_"min"=25/1` |
|
| 143. |
Two sources with intensity `I_(0)` and `4I_(0)` respectively interefere at a point in a medium. Then the maximum and minimum possible intensity would be |
|
Answer» `I_(max)=(sqrt(l_(1))+sqrt(l_(2)))^(2)` Here, `l_(1)=l_(0)` and `l_(2)=4I_(0)` `:. I_(max)=(sqrt(l_(0))+sqrt(4l_(0)))^(2)=9I_(0)` and `l_("min")=(sqrt(l_(1))-sqrt(l_(2)))^(2)` `=l_(0)` |
|
| 144. |
In normal YDSE experiment maximum intensity is `4I_(0)` in Column I, y-coordinate is given corresponding to centre line In Coloum II resultant intensities are given Match the two columns. `{:(,"Column I",,"Column II"),("A",y=(lambdaD)/d,,p. I=I_(0)),("B",y=(lambdaD)/(2d),,q. I=2I_(0)),("C",y=(lambdaD)/(3d),,r. I=4I_(0)),("D",y=(lambdaD)/(4d),,s. I=zero):}` |
|
Answer» Correct Answer - A::B::C::D `Deltax=(yd)/D,phi=(2pi)/lambda.Deltax and I=4I_(0)"cos"^(2)phi/2` |
|
| 145. |
In a YDSE, the separation between slits is `2 mm` where as the distance of screen from the plane of slits is `2.5 m`. Light of wavelengths in the range `200-800 nm` is allowed to fall on the slits. Find the wavelengths in the visible region that will be present on the screen at `1 mm` from central maximum. Also find the wavelength that will be present at that point of screen in the infrared as well as in the ultraviolet region.A. `4000Å`B. `5000Å`C. `6000Å`D. `8000Å` |
|
Answer» Correct Answer - D |
|
| 146. |
Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratioA. (a) `25:1`B. (b) `5:1`C. (c) `9:4`D. (d) `25:16` |
|
Answer» Correct Answer - C `(I_(max))/(I_(min))=((sqrt(I_1/I_2)+1)/(sqrt(I_1/I_2)-1))^2impliesI_1/I_2=9/4` |
|
| 147. |
The polarising angle for a medium is `60^(@)`. Determine (i) the refraction index of the medium and (ii) the refracting angle. |
|
Answer» Correct Answer - (i) 1.732 (ii) `30^(@)` |
|
| 148. |
Two monochromatic coherent point sources `S_(1)` and `S_(2)` are separated by a distance L. Each sources emits light of wavelength `lambda`, where `L gt gt lambda`. The line `S_(1) S_(2)` when extended meets a screen perpendicular to it at point A. ThenA. the interference fringe on the screen are straight lines shapeB. the interference fringes on the screen are strainght lines perpendicular to the line `S_(1) S_(2) A`C. point A is an intensity maxima if `L = n lambda`D. point A is always an intensity maxima for any separation L |
|
Answer» Correct Answer - c If the screen is prependicular to y-axis (line joinin the sources), i.e., xz plane, the fringes will be circular (i.e., concentric circles with their centers on the point of intersection of the screen with y-axis). In this situation, central fringe will be bright id `S_(1), S_(2) = n lambda` and dark if `S_(1) S_(2) = (2 n - 1) lambda // 2`. From all this, it is clear that shape of fringes depends on the nature of sources and direction of observation, i.e., position of screen |
|
| 149. |
If the polarising angle for air glass interface is `56.3^(@)`, what is the angle of refraction in glass ? |
| Answer» `becausei_(p)+r_(p)=90^(@)" " thereforer_(p)=90^(@)-i_(p)=90^(@)-56.3^(@)=33.7^(@)` | |
| 150. |
Light waves form two coherent source having internsity ration `81:1` produce interference. Then, the ratio of maxima and minima in the interference pattern will be |
|
Answer» Given, `(I_(1))/(I_(2))=(A_(1)^(2))/(A_(2)^(2))=(81)/(1)` `:. (A_(1))/(A_(2))=(9)/(1)` or `A_(1)=9 A_(2)` …….(1) `:. (I_(max))/(I_(min))=((A_(1)+A_(2))^(2))/((A_(1)-A_(2))^(2))` " From Eq. " (i), we get `(I_(max))/(I_("min"))= ((9A_(2)+A_(2))^(2))/((9A_(2)-A_(2))^(2)) = ((10)^(2))/((8)^(2)) =(25)/(16)` |
|