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.
| 401. |
This interference film is used to measure the thickness of slides, paper, etc. The arrangement is as shown in fig. For the sake of clarity, the two strips are shown thick. Consider the wedge formed in between strips 1 and 2. If the interference pattern because of the two waves reflected from wedge surface is observed, then from the observed data we can compute thickness of paper, refractive index of the medium filled in wedge, number of bonds formed, etc. Considre the strips to be thick as compared to wavelength of ligth and light is incident normally. Neglect the effect due to reflection from top surface of strip 1 and bottom surface of strip 2. Take `L = 5 cm` and `lambda_(air) = 40 nm`. For strip 1 refractive index is 1.34 abd for strip 2 refractive index is 1.6 The wedge. is filled with a medium having refractive index 1.5. ThenA. the band jat contact point would be darkB. the band at contact point would be brightC. at contact point, maxima or minima occursD. at contact point, uniform illumination would be there |
|
Answer» Correct Answer - b For reflection from bottom layer of strip 1 as well as for top surface of strip 2, phase shift to `pi` takes place. So, conditions for maxima is `2 t = n lambda_("medium")` and condition for minima is `2 t = (2n - 1) lambda_("medium") //2`. So, contact would be bright as here path difference is 0. |
|
| 402. |
This interference film is used to measure the thickness of slides, paper, etc. The arrangement is as shown in fig. For the sake of clarity, the two strips are shown thick. Consider the wedge formed in between strips 1 and 2. If the interference pattern because of the two waves reflected from wedge surface is observed, then from the observed data we can compute thickness of paper, refractive index of the medium filled in wedge, number of bonds formed, etc. Considre the strips to be thick as compared to wavelength of ligth and light is incident normally. Neglect the effect due to reflection from top surface of strip 1 and bottom surface of strip 2. Take `L = 5 cm` and `lambda_("air") = 40 nm`. In question 53, if air wedge has been filled with a medium having refractive index 1.3 then find the number of bright bands.A. 199B. 99C. 499D. 130 |
|
Answer» Correct Answer - d Here `lambda_("air")` has to replaced by `lambda_("medium")` `lambda_("medium") = (lambda_("air"))/(mu) = (400)/(1.3) nm` For bright band, `2 t =(2n - 1) (lambda_("meduim"))/(2)` `(2 h x)/(L) = (2n - 1) (lambda_("meduim"))/(2)` `x = (2n - 1) [(400 xx 10^(-9) xx 5 xx 10^(-2))/(2 xx 1.3 xx 2 xx 20 xx10^(-6))]` For maximum `n, x = L` `n_(max)` `= 130.5` |
|
| 403. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. Let `DeltaX_(A)` and `DeltaX_(C)` represent path differences between waves interering at 1 and 3 respectively then `(|DeltaX_(C)|-(|DeltaX_(A)|)` is equal to |
|
Answer» Correct Answer - B `CBFimplies` path difference `=0` `DeltaX_(A)toAt" "1` dark fringe having path difference `=(lamda)/(2)` `DeltaX_(C)to`At 3, bright fringes path difference `=lamda` `therefore(|DeltaX_(C)|-|DeltaX_(A)|)=(lamda-(lamda)/(2))=(lamda)/(2)=(600)/(2)nm=300nm` |
|
| 404. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. Let `DeltaX_(A)` and `DeltaX_(C)` represent path differences between waves interering at 1 and 3 respectively then `(|DeltaX_(C)|-(|DeltaX_(A)|)` is equal to |
|
Answer» Correct Answer - b `Delta X_(C) = lambda, Delta X_(A) = (lambda)/(2)` `Delta X_(C) - Delta X_(A) = (lambda)/(2) = 3000` nm |
|
| 405. |
The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm. Q. The third order bright fringe isA. 2B. 3C. 4D. 5 |
|
Answer» Correct Answer - d Order of the fringe can be counted on either side of the central maximum. For example, Fringe no.3 is first-order brigth fringe. |
|
| 406. |
Light of wavelength 6000Å is incident on a slit of width 0.30 mm. The screen is placed 2 m from the slit. Find (a) the position of the first dark fringe and (b). The width of the central bright fringe. |
|
Answer» The first fringe is on either side of the central bright fringe. here `n=+-1,D=2m,lamda=6000Å=6xx10^(-7)m` `thereforesintheta=(x)/(D)impliesa=0.30mm=3xx10^(-4)m` `impliesasintheta=nlamdaimplies(ax)/(D)=nlamda` (a). `x=(nlamdaD)/(a)impliesx=+-[(1xx6xx10^(-7)xx2)/(3xx10^(-4))]=+-4xx10^(-3)m` The positive and negative signs corresponds to the dark fringes on either side of the central bright fringe. (b). The width of the central bright fringe `y=2x=2xx4xx10^(-3)=8xx10^(-3)m=8mm` |
|
| 407. |
An interference is observed due to two coherent sources `S_(1)` placed at origin and `S_(2)` placed at (0, 3l, 0). Here, `lambda` is the wavelength of the sources. Both sources are having equal intensity `I_(0)`. A detector D is moved along the positive x-axix. Find x-coordinates on the x-axis (exchange `x = 0` and `x = OO`).A. `x = 4 lambda`B. `x = 7 lambda//4`C. `x = 5 lambda//4`D. `x = 3 lambda` |
|
Answer» Correct Answer - a.,c At ` x = 0`, path difference is `3 lambda`. Hence,third-order maximum will be obtained. At `x = oo`, path difference is zero. Hence, zero-order maxima is obtained. In between, first - and second-order maxima will be obtained. First order maxima: `S_(2) P - S_(1) P = lambda` or `sqrt(x^(2) + 9lambda^(2)) - x = lambda` or `sqrt(x^(2) + 9 lambda^(2)) = x + lambda` Squaring this, we get `x^(2) + 9 lambda^(2) = x^(2) + lambda^(2) + 2 x lambda` Solving this, we get `x = 4 lambda` Second-order maxima: `S_(2) P - S_(1) P = 2 lambda` or `sqrt(x^(2) + 9 lambda^(2)) - x = 2 lambda` or ` sqrt( x^(2) + 9 lambda^(2)) = (x + 2 lambda)` Squaring both sides, we get `x^(2) + 9 lambda^(2) = x^(2) + 4 lambda^(2) + 4 x lambda` Solving this, we get `x = (5)/(4) lambda = 1.25 lambda` Hence, the desired x-corrdinates are `x = 1.25 lambda` and `x = 4 lambda` |
|
| 408. |
In the arrangment shown in fin. For what minimum value of d is there a dark bant at point O on the screen?A. `sqrt((D lambda)/(4))`B. `sqrt((3D lambda)/(4))`C. `sqrt((D lambda)/(8))`D. `sqrt((2D lambda)/(43)` |
|
Answer» Correct Answer - a For dark spot at O, `Delta x = (AB + BO) - AO` `= 2(D^(2) + d^(2))^(1//2) - 2 D = 2D [(1 + (d^(2))/(D^(2)))^(1//2)-1]` `implies Delta x =(d^(2))/(D)` `(d^(2))/(D) = (lambda)/(2) implies d = sqrt((D lambda)/(4)` `Detla x` at P: `Delta x = (AB + BP) - (AC + PC)` `= (AB - AC) + (BP - PC)` `Delta x` at O: `Delta x = (lambda)/(2)` The path difference at P will be zero if `x = d`. Hence next maxima will front at P. Fringe width = distance between two consecutive dark to brigth fringes `= 2 xx` distance between two consective bright dark fringes `= 2 xx d` |
|
| 409. |
An interference is observed due to two coherent sources `S_(1)` placed at origin and `S_(2)` placed at (0, 3l, 0). Here, `lambda` is the wavelength of the sources. Both sources are having equal intensity `I_(0)`. A detector D is moved along the positive x-axix. Total number of black points observed by the observer on positive x-axis isA. twoB. fourC. threeD. five |
|
Answer» Correct Answer - b At ` x = 0`, path difference is `3 lambda`. Hence,third-order maximum will be obtained. At `x = oo`, path difference is zero. Hence, zero-order maxima is obtained. In between, first - and second-order maxima will be obtained. First order maxima: `S_(2) P - S_(1) P = lambda` or `sqrt(x^(2) + 9lambda^(2)) - x = lambda` or `sqrt(x^(2) + 9 lambda^(2)) = x + lambda` Squaring this, we get `x^(2) + 9 lambda^(2) = x^(2) + lambda^(2) + 2 x lambda` Solving this, we get `x = 4 lambda` Second-order maxima: `S_(2) P - S_(1) P = 2 lambda` or `sqrt(x^(2) + 9 lambda^(2)) - x = 2 lambda` or ` sqrt( x^(2) + 9 lambda^(2)) = (x + 2 lambda)` Squaring both sides, we get `x^(2) + 9 lambda^(2) = x^(2) + 4 lambda^(2) + 4 x lambda` Solving this, we get `x = (5)/(4) lambda = 1.25 lambda` Hence, the desired x-corrdinates are `x = 1.25 lambda` and `x = 4 lambda` |
|
| 410. |
In the arrangment shown in fin. Find the fringe width.A. dB. 2dC. 4dD. 3d |
|
Answer» Correct Answer - b For dark spot at O, `Delta x = (AB + BO) - AO` `= 2(D^(2) + d^(2))^(1//2) - 2 D = 2D [(1 + (d^(2))/(D^(2)))^(1//2)-1]` `implies Delta x =(d^(2))/(D)` `(d^(2))/(D) = (lambda)/(2) implies d = sqrt((D lambda)/(4)` `Detla x` at P: `Delta x = (AB + BP) - (AC + PC)` `= (AB - AC) + (BP - PC)` `Delta x` at O: `Delta x = (lambda)/(2)` The path difference at P will be zero if `x = d`. Hence next maxima will front at P. Fringe width = distance between two consecutive dark to brigth fringes `= 2 xx` distance between two consective bright dark fringes `= 2 xx d` |
|
| 411. |
Consider the situation shown in fig. The two slits `S_(1)` and `S_(2)` placed symmetrically around the central line are illuminated by monochromatic light of wavelength `lambda`. The separation between the slit is d. The ligth transmitted by the slits falls on a screen `S_(0)` placed at a distance D form the slits. The slit `S_(3)` is at the central line and the slit `S_(4)` is at a distance z from `S_(3)` Another screen `S_(c)` is placed a further distance D away from `S_(c)` Find the ratio of the maximum to minimum intensity observed on `S_(c)` If `z = (lambda D)/(d)`A. 4B. 2C. `oo`D. 1 |
|
Answer» Correct Answer - c `z = (lambda d)/(d)` `Delta x` at `S_(4): Delta x = (lambda D)/(d) (d)/(D) = lambda` Hence, maxima at `S_(4)` as well as `S_(3)` Resultant intensity at `S_(4). I = 4 I_(0)` `:. (I_(max))/(I_(min)) = ([(4 I_(0))^(1//2) + (4 I_(0))^(1//2)]^(2))/([(4 I_(0))^(1//2) - (4 I_(0))^(1//2)]^(2) )= oo` |
|
| 412. |
Consider the situation shown in fig. The two slits `S_(1)` and `S_(2)` placed symmetrically around the central line are illuminated by monochromatic light of wavelength `lambda`. The separation between the slit is d. The ligth transmitted by the slits falls on a screen `S_(0)` placed at a distance D form the slits. The slit `S_(3)` is at the central line and the slit `S_(4)` is at a distance z from `S_(3)` Another screen `S_(c)` is placed a further distance D away from `S_(c)` Find the ratio of the maximum to minimum intensity observed on `S_(c)` If `z = (lambda D)/(4 d)`A. `[3 - 2 sqrt 2]^(2)`B. `[3 + sqrt 2]^(2)`C. `[3 - sqrt 2]^(2)`D. `[3 + 2 sqrt 2]^(2)` |
|
Answer» Correct Answer - d `z = (lambda D)/(4d)` `Delta x = y (d)/(D) = (lambda D)/(4 d) (d)/(D) = (lambda)/(4)` `phi = (2 pi)/(lambda) ((lambda)/(4)) = (pi)/(4)` Intensity at `S_(4): I_(4) = 2 I_(0) (1 + cos ((pi)/(2))) = 2 I_(0)` Intensity at `S_(3): I_(3) = 4 I_(0)` `:. (I_(max))/(I_(min))=([(4I_(0))^(1//2)+(2I_(0))^(1//2)]^(2))/([(4I_(0))^(1//2)-(2I_(0))^(1//2)]^(2))=[((2+sqrt(2)))/((2-sqrt(2)))]^(2)` `= [((2 + sqrt 2)^(2))/(4 - 2)]^(2) = (1)/(4) [4 + 2 + 4 sqrt 2]^(2)` `= (1)/(4) [ 6 + 4 sqrt 2]^(2) = [3 + 2 sqrt 2]^(2)` |
|
| 413. |
Consider the situation shown in fig. The two slits `S_(1)` and `S_(2)` placed symmetrically around the central line are illuminated by monochromatic light of wavelength `lambda`. The separation between the slit is d. The ligth transmitted by the slits falls on a screen `S_(0)` placed at a distance D form the slits. The slit `S_(3)` is at the central line and the slit `S_(4)` is at a distance z from `S_(3)` Another screen `S_(c)` is placed a further distance D away from `S_(c)` Find the ratio of the maximum to minimum intensity observed on `S_(c)` If `z = (lambda D)/(2 d)`A. 1B. `1//2`C. `3//2`D. 2 |
|
Answer» Correct Answer - a `z = (lambda D)/(2 d)` At `S_(4): (Delta x)/(4) = (z)/(D)` `implies Delta x = (lambda D)/(2 d) = (lambda)/(2)` Hence, minima at `S_(4)` and maxima at `S_(3)` (intensity `4 I_(0)`). Hence, `(I_(max))/(I_(min)) = 1` |
|
| 414. |
The critical angle of a for a medium is `45^(@)`. What is its polarising angle. |
|
Answer» Correct Answer - `54.7^(@)` |
|
| 415. |
An electromagnetic wave travelling through a transparent medium is given by `E_(x)(y,t)=E_(ax)sin2pi[y/(5xx10^(-7))-3xx10^(14)t]` in SI unit then the refractive index of medium is, |
| Answer» Correct Answer - 2 | |
| 416. |
Central fringe obtained in diffraction pattern due to a single slit-A. is of minimum intensityB. is of maximum intensityC. intensity does not depend upon slit widthD. none of the above |
| Answer» Correct Answer - B | |
| 417. |
In a diffraction pattern the width of any fringe isA. directly proportional to slit widthB. inversely proportional to slit widthC. Independent of the slit widthD. None of the above |
| Answer» Correct Answer - B | |
| 418. |
A diffraction pattern is obtained using a beam of redlight. What happens if the red light is replaced by blue lightA. no changeB. diffraction bands become narrower and crowded togetherC. bands become broader and farther apartD. bands disappear |
| Answer» Correct Answer - B | |
| 419. |
Polaroid are usedA. to eliminate head light glare in automobileB. in production of 3-D motion picturesC. in sun glassesD. all the above |
| Answer» Correct Answer - D | |
| 420. |
When light falls on two polaroid sheets, one observes complex brightness then the two polaroids axes areA. Mutually perpendicularB. Mutually parallelC. Angle between their two axes is `45^(@)`D. None of the above |
| Answer» Correct Answer - B | |
| 421. |
A beam of ordinary light is incident on a system of four polaroids which are arranged in succession such that each polaroid is turned through `30^(@)` with respect to the preceding one. The percentage of the incident intensity that emerges out from the system is appromatelyA. `56%`B. `6.25%`C. `21%`D. `14%` |
|
Answer» Correct Answer - C `I = (I_(0))/(2)(cos^(2)theta)^(n-1)` where n = number of polaroids |
|
| 422. |
In a double-slit experiment, the slits are separated by a distance d and the screen is at a distance D from the slits. If a maximum is formed just opposite to each slit, then what is the order or the fringe so formed?A. `(d^(2))/(2 lambda D)`B. `(2 d^(2))/( lambda D)`C. `(d^(2))/( lambda D)`D. `(d^(2))/(4 lambda D)` |
|
Answer» Correct Answer - a `y_(n) (max) = n(D lambda)/(d)` Hence, `y_(n) (max) = d // 2` So, `n (d lambda)/(d) = (d)/(2)` or `n = (d^(2))/(2 lambda D)` |
|
| 423. |
A parallel beam of white light is incident on a thin film of air of uniform thickness. Wavlength `7200 Å` and `5400 Å` are observed to be missing from the spectrum of reflected light viewed normally. The other wavelength in the visible region missing in the reflected spectrum isA. `6000 Å`B. `4320 Å`C. `5500 Å`D. `6500 Å` |
|
Answer» Correct Answer - b The wavelength missing from the reflected spectrum must satisfy the condition,`2 mu t = n lambda`, where t is thickness of air film. `2 mu t = n lambda_(1) = (n + 1) lambda_(2)` or `n xx (7200) = (n + 1) 5400` `:.n = 3` The next wavelengths must satisfy the condidtion, `n lambda_(1) = (n + 2) lambda_(2)` or `7200 xx 3 = (3 + 2) lambda_(2) = 5 lambda_(2)` `implies lambda_(2) = 4320 Å`. |
|
| 424. |
A plane electromagnetic wave of frequency `w_(0)` falls normally on the surface of a mirror approaching with a relativisitic velocity v. Then frequency of the reflected wave will be (given `beta=(v)/(c )`):A. `((1-beta)/(1+beta))w_(0)`B. `(1+beta)/((1-beta)w_(0))`C. `((1+beta)w_(0))/((1-beta))`D. `((1-beta))/((1+beta)w_(0))` |
|
Answer» Correct Answer - A |
|
| 425. |
Ratio of intensities between a point A and that of central fringe is 0.853. If we consider slits are of equal width then path difference between two waves at point A will be :A. `(lambda)/(2)`B. `(lambda)/(4)`C. `(lambda)/(8)`D. `lambda` |
|
Answer» Correct Answer - C |
|
| 426. |
One radio transmitter A operating `60.0MHz`is `10.0`from another similar transmitter `B` that is `180^(@)` out of phase with transmitter`A`.How far must be observer move from transmitter A towards transmitter B along the line connencting A andB to reach the nearest point when the two beam are in phase? |
|
Answer» Correct Answer - [1.25m] |
|
| 427. |
Light propagates rectilinearly, due toA. wave natureB. wavelengthsC. velocityD. frequency |
|
Answer» Correct Answer - A Light propagates rectilinearly due to wave nature. |
|
| 428. |
The refractive index of glass is 1.5 for light waves of `lamda=6000` Ã… in vacuum. Its wavelength in glass isA. 2000 Ã…B. 4000 Ã…C. 1000 Ã…D. 3000 Ã… |
|
Answer» Correct Answer - B `mu=(c)/(v)=(lambda_(v))/(lambda_(g))` ` :. lambda_(g)=(lambda_(v))/(mu)=(6000)/(1.5) = 4000Ã…` |
|
| 429. |
The wavelength range of the light that is visible to an average human being is 400 nm to 700 nm . What is the frequency range of this visile light ? |
|
Answer» Correct Answer - `4.3 xx 10^(14)` Hz to `7.5 xx 10^(14)` Hz |
|
| 430. |
Two identical monochromaticlight sources A & B intensity `10^(-15)W//m^(2)` produce wavelength of light `4000sqrt(3)Å` A glass of thickness 3mm is placed in the path of the ray as shown in figure the glass has a varible refractive index `n=1+sqrt(x)` where x (in mm) is distance of plate from left to right calculate total intensity at focal points F of the lens. |
|
Answer» Correct Answer - `[4xx10^(-15)W//m^(2)]` |
|
| 431. |
The wavelength of red light in air is 7890 Å . What is the wavelength in glass `(mu = 1.5)` ? |
|
Answer» Correct Answer - 5260 Å |
|
| 432. |
If A is the amplitude of the wave coming from a line source at a distance r, then :A. `A prop r^(-2)`B. `A prop r^(-1)`C. `A prop r^(2)`D. `A prop r^(1)` |
|
Answer» Correct Answer - B |
|
| 433. |
The two waves represented by `y_1=a sin (omegat)` and `y_2=b cos (omegat)` have a phase difference ofA. (a) `0`B. (b) `pi/2`C. (c) `pi`D. (d) `pi/4` |
|
Answer» Correct Answer - B `y_1=a sin omegat` and `y_2=b cos omega t=bsin (omegat+pi/2)` So phase difference `varphi=pi//2` |
|
| 434. |
In a wave, the path difference corresponding to a phase difference of `phi` isA. (a) `(pi)/(2lambda)varphi`B. (b) `pi/lambdavarphi`C. (c) `(lambda)/(2pi)varphi`D. (d) `(lambda)/(pi)varphi` |
|
Answer» Correct Answer - C For `2pi` phase difference `rarr` Path difference is `lambda` `:.` For `phi` phase difference `rarr` Path difference is `(lambda)/(2pi)xxvarphi` |
|
| 435. |
Two waves are represented by the equations `y_1=a sin omega t` and `y_2=a cos omegat`. The first waveA. (a) Leads the second by `pi`B. (b) Lags the second by `pi`C. (c) Leads the second by `pi/2`D. (d) Lags the second by `pi/2` |
|
Answer» Correct Answer - D `y_1=a sin omegat`, `y_2=a cos omegat=a sin (omegat+pi/2)` |
|
| 436. |
The wavelength of light coming from a sodium source is 589 nm. What will be its wavelength in water? Refractive index of water 1.33.` |
|
Answer» The wavelength in water is `lambda = lambda_(0) // mu,` where `lambda_(0)` is the wavelength in vaccum and `mu` is the refractive index of water. Thus `lambda = (589)/(1.33) = 443` nm, |
|
| 437. |
A monochromatic light of `lambda = 500 Å` is incident on two indentical slits separated by a distance of `5 xx 10^(-4)` m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness `1.5 xx 10^(-6)` m and refractive index `mu = 1.5` is placed between one of the slits and the screen . Find the intensity at the center of the screen. |
|
Answer» In case of interference `I_(R) = I_(1) + I_(2) + 2 (sqrt (I_(1) I_(2))) cos phi` Now, as for identical slits `I_(1) = I_(2) = I,` so `I_(R) = 2 (1 + cos phi) = 4 I cos^(2) (phi // 2)` But for central maxima, `phi = 0^(@),` and here `I_(R) = I_(0)` (given) `I_(0) = 4 I cos (0^(@)) = 4 I` Hence, `I_(R) = I_(0) cos^(2) ((phi)/(2))` Now, when the glass plate is introduced, path difference between the waves at the position of central maxima will become `Delta x = ( mu - 1) t = (1.5 - 1) 1.5 xx 10^(-6) = 7.5 xx 10^(-7) m` `:. phi = (2 pi)/(lambda) (Delta x) = (2 pi)/(5 xx 10^(-7)) xx 7.5 xx 10^(-7) = 3 pi` So, intensity at central maxima will now be `I_(R) = I _(0) cos^(2) ((3 pi)/(2)) = I_(0) xx 0 = 0` Also, from theory of intergerence, fringe shift `y_(0) = (D)/(d) (mu - 1) t` Which in the light of Eq. (ii) and given data becomes `y_(0) = 7.5 xx 10^(-7) = 1.5 xx 10^(-3) m = 1.5 mm`. |
|
| 438. |
The bending of light about corners of an obstacle is calledA. DispersionB. RefractionC. DeviationD. Diffraction |
| Answer» Correct Answer - D | |
| 439. |
The bending of light about corners of an obstacle is calledA. ReflectionB. DiffractionC. RefractionD. Interference |
| Answer» Correct Answer - B | |
| 440. |
A solution of camphor in alcohol in a tube 20cm long is found to effect a rotation of the plane of viberation of light passing is of `33^(@)`. What must be the indentify of camphor in `g//cm^(2)` in solution? The specific rotation of camphor is `+54^(@)`. |
|
Answer» Correct Answer - `0.3055xx gcm^(-3)` |
|
| 441. |
A `20cm` length of a certain solution causes right-handed rotation of `38^@`. A `30cm` length of another solution causes left-handed rotation of `24^@`. The optical rotation caused by `30cm` length of a mixture of the above solutions in the volume ratio `1:2` isA. (a) Left handed rotation of `14^@`B. (b) Right handed rotation of `14^@`C. (c) Left handed rotation of `3^@`D. (d) Right handed rotation of `3^@` |
|
Answer» Correct Answer - D As `thetaproplambda` Volume ratio `1:2` in a tube of length `30cm` means `10cm` length of first solution and `20cm` length of second solution. Rotation produced by `10cm` length of first solution `theta_1=(38^@)/(20)xx10=19^@` Rotation produced by `20cm` length of second solution `theta_2=(24^@)/(30)xx20=-16^@` `:.` Total rotation produced `=19^@-16^@=3^@` |
|
| 442. |
Given: width of aperture `d=3mm` and `lambda=500nm`. For what distance ray optics is a good approximation?A. (a) `18m`B. (b) `18nm`C. (c) `18Å`D. (d) 18 light years |
|
Answer» Correct Answer - A `D_F=d^2/lambda` `D_F=((3xx10^-3)(3xx10^-3))/(500xx10^-9)m` `=90/5m=18m` |
|
| 443. |
A `20cm` length of a certain solution causes right-handed rotation of `38^@`. A `30cm` length of another solution causes left-handed rotation of `24^@`. The optical rotation caused by `30cm` length of a mixture of the above solutions in the volume ratio `1:2` isA. Left handed rotation of `14^(@)`B. Right handed rotation of `14^(@)`C. Left handed rotation of `3^(@)`D. Right handed rotation of `3^(@)` |
|
Answer» Correct Answer - A |
|
| 444. |
A polariser in used toA. (a) Reduce intensity of lightB. (b) Produce polarised lightC. (c) Increase intensity of lightD. (d) Produce unpolarised light |
|
Answer» Correct Answer - B Polariser produced polarised light. |
|
| 445. |
Light waves can be polarised as they areA. (a) TransverseB. (b) Of high frequencyC. (c) LongitudinalD. (d) Reflected |
|
Answer» Correct Answer - A Only transverse waves can be polarised. |
|
| 446. |
Which of the following is incorrect?A. it the wave is longitudinal then it must be a mechanical waveB. if the wave is mechanical then it may or not be transverse waveC. Mechanical waves cannot propagate in vacuumD. Diffractor helps us to distinguish between sound wave and light wave |
|
Answer» Correct Answer - D Polarisation help us to distinguish between sound wave and light wave |
|
| 447. |
An optically active compoundA. rotates the plane of polarised lightB. changes the direction of polarised lightC. does not allow plane polarised light to pass throughD. none of these. |
|
Answer» Correct Answer - A An optically active compound rotates the plane of polarized light. |
|
| 448. |
An optically active compoundA. (a) Rotates the plane polarised lightB. (b) Changing the direction of polarised lightC. (c) Do not allow plane polarised light to pass throughD. (d) None of the above |
|
Answer» Correct Answer - A When the plane-polarised light passes through certain substance, the plane of polarisation of the light is rotated about the direction of propagation of light through a certain angle. |
|
| 449. |
An optically active compoundA. Rotates the plane polarised lightB. Changing the direction of polarised lightC. Do not allow plane polarised light to pass throughD. None of the above |
|
Answer» Correct Answer - C |
|
| 450. |
An optically active compoundA. rotates the plane polarised lightB. changes the direction of polarised lightC. does not allow plane polarised light to pass throughD. None of the above |
|
Answer» Correct Answer - A When the plane polarised light passes through certain substance, the plane of polarisation of the light is rotated about the direction of propagation of light through a certain angle |
|