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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.
651. |
Specific rotation of sugar solution is 0.5 deg `m^(2)//kg`. `200kgm^(-3)` of impure sugar solution is taken in a sample polarimeter tube of length 20 cm and optical rotation is found to be `19^(@)`. The percentage of purity of sugar is :A. 0.2B. 0.8C. 0.95D. 0.89 |
Answer» Correct Answer - D |
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652. |
Monochromatic green light of wavelength 550 nm illuminates two parallel narrow slits `7.7mu m` apart. The angular deviation `theta` of third order (for m =3) bright fringe in radian and in degreeA. `21.6, 12.4^(@)`B. `0.216, 1.24^(@)`C. `0.216, 12.4^(@)`D. `216, 1.24^(@)` |
Answer» Correct Answer - C `theta = (lambda)/(d)` |
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653. |
Monochromatic green light of wavelength `5 xx 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 isA. 0.25mmB. 0.1mmC. 1.0mmD. 0.01mm |
Answer» Correct Answer - C | |
654. |
Monochromatic green light of wavelength `5 xx 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 isA. `0.25` mmB. `0.1 mm`C. `1.0 mm`D. `0.01mm` |
Answer» Correct Answer - C Separation of bright lines or fringe width `beta=(Dlambda)/d=(5xx10^(-7)xx2)/10^(-3)=10^(-3)m=1.0mm` |
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655. |
Assertion : The increase in wavelength due to doppler effect is termed as red shift. Reason : In red shift, a wavelength in the middle of the visible region of the spectrum moves towards the violet end of the spectrum.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
Answer» Correct Answer - C Astronomers call the increase in wavelength due to doppler effect as red shift since wavelength in the middle of the visible region of the spectrum moves towards the red end of the spectrum. |
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656. |
Doppler shift for the light of wavelength `6000Å` emitted from the sun is `0.04 Å`. If radius of the sun is `7 xx 10^(8)m` then time period of rotation of the sun will be :A. 30 daysB. 365 daysC. 24 hoursD. 25 days |
Answer» Correct Answer - D |
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657. |
A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength `550 Å` is :A. `616 Å`B. `1833Å`C. `5500Å`D. `6160Å` |
Answer» Correct Answer - B |
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658. |
A star is moving away from an observer with a speed of `500 kms^(-1)` .Calculate the Doppler shift if the wavelength of light emitted by the star is 6000`Å`. |
Answer» Correct Answer - Increase of `10Å` |
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659. |
Light emitted from a distant star is observed at frequency 5000 MHz for the star stationary with respect to us. If the star starts approaching us with speed `6xx10^(5)m//s`. Then the observed frequency will be |
Answer» `(Deltaf)/f=v/c` `Deltaf=(5xx10^9xx6xx10^5)/(3xx10^8)=10xx10^6` =`10^7` Hz = 10 MHz As the source is approaching the observer Observed frequency =`f+Deltaf`=5010 MHz |
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660. |
An astronaut approaching the moon sends a radio signal of frequency `5 xx 10^(3)` MHz towards moon to find the speed of his rocket . The frequency of waves reflected back from the moon to find the speed of his rocket . The frequency .Calculate the velocity of the rocket relative to the moon. |
Answer» Correct Answer - `2.58 km s^(-1)` |
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661. |
STATEMENT-1 The two silts in `YDSE` are illuminated by two different sodium lamps emitting light of same wavelength .No interference pattern will be observed. STATEMENT-2 Two independent light sources (except `LASER`)cannot be coherent.A. Statement-1 is true Statemetnt-2 is True,Statement -2 is a correct explanation for statement -1.B. Statement-1 is true Statemetnt-2 is True,Statement -2 is NOT a correct explanation for statement -1.C. Statement -1 is true,statement -2 is falseD. Statement -1 is False ,statement -2 is True. |
Answer» Correct Answer - A | |
662. |
Two coherent monochromatic sources A and B emit light of wavelength `lambda`. The distance beween A and B is `d = 4lambda`. (a) If a light detector is moved along a line CD parallel to AB, what is the maximum number of minima observed ? (b) If the detector is moved along a line BE perpendicular to AB and passing through B, what is the number of maxima observed ? |
Answer» Correct Answer - [9,9] |
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663. |
Two coherent monochromatic sources A and B emit light of wavelength `lambda`. The distance beween A and B is `d = 4lambda`. (a) If a light detector is moved along a line CD parallel to AB, what is the maximum number of minima observed ? (b) If the detector is moved along a line BE perpendicular to AB and passing through B, what is the number of maxima observed ? |
Answer» Correct Answer - [(a) 8, (b) 4] |
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664. |
STATEMENT-1 Superposition takes place only between those waves emitted by coherent sources. STATEMENT-2 All coherent sources emit energy in proper order.A. Statement-1 is true Statemetnt-2 is True,Statement -2 is a correct explanation for statement -1.B. Statement-1 is true Statemetnt-2 is True,Statement -2 is NOT a correct explanation for statement -1.C. Statement -1 is true,statement -2 is falseD. Statement -1 is False ,statement -2 is True. |
Answer» Correct Answer - D | |
665. |
Two point sources X and Y emit waves of same frequency and speed but Y lags in phase behind X by `2pi//radian`. If there is a maximum in direction D the distance XO using n as an integer is given by A. `(lamda)/(2)(n-1)`B. `lamda(n+1)`C. `(lamda)/(2)(n+l)`D. `lamda(n -l)` |
Answer» Correct Answer - B |
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666. |
Two point sources X and Y emit waves of same frequency and speed but Y lags in phase behind X by `2pi//radian`. If there is a maximum in direction D the distance XO using n as an integer is given by A. (a) `lambda/2(n-l)`B. (b) `lambda(n+1)`C. (c) `lambda/2(n+1)`D. (d) `lambda(n-1)` |
Answer» Correct Answer - B For maxima `2pin=(2pi)/(lambda)(XO)-2pil` or `(2pi)/(lambda)(XO)=2pi(n+l)` or `(XO)=lambda(n+l)` |
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667. |
A 300 mm long tube containing 60 `cm^(3)` of sugar solution produces an optical rotations of `10^(@)` when placed in a saccharimeter. If specific rotation of sugar is `60^(@)` calculate the quantity of sugar constained in te tube solution. |
Answer» `l=300mm=30cm=3` decimeter, `theta=10^(@),[alpha]_(A)^(T)=60^(@)`, volume of solution `=60cm^(3)` `theta=[alpha]_(lamda)^(T)lCimpliesC=(theta)/([alpha]_(lamda)^(T)l)=(10^(@))/(60^(@)xx3)=(1)/(18)gcm^(-3)` Quantity of sugar contained `=(1)/(18)xx60=3.33g` |
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668. |
A calcite crystal is placed over a dot on a piece of paper and rotated, on seeing through the calcite one will be seeA. (a) One dotB. (b) Two stationary dotsC. (c) Two rotating dotsD. (d) One dot rotating about the other |
Answer» Correct Answer - D Light suffers double refraction through calcite. |
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669. |
When an unpolarized light of intensity `I_(0)` is incident on a polarizing sheet, the intensity of the light which does not get transmitted isA. zeroB. `I_(0)`C. `(1)/(2)I_(0)`D. `(1)/(4)I_(0)` |
Answer» Correct Answer - C `I=I_(0)cos^(2)theta` Intensity of polarized light `=(I_(0))/(2)` `therefore` Intensity of untransmitted light `=I_(0)-(I_(0))/(2)=(I_(0))/(2)` |
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670. |
A light is incident on a transparent medium of `mu = 1.732` at the polarising angle. The angle of reaction isA. `60^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
Answer» Correct Answer - B `mu = Tan i_(p) I_(p)+r=90` |
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671. |
A beam of electron is used `YDSE` experiment . The slit width is d when the velocity of electron is increased ,thenA. (a) no interference is observedB. (b) fringe with increasesC. (c) fringe width decreasesD. (d) fringe width remains same |
Answer» Correct Answer - B Momentum of the electron will increase. So the wavelength `(lambda=h//p)` of electron will decrease and fringe width decreases as `betaproplambda` |
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672. |
Statement I: Fringe width depends upon refractive index of the medium. Statement II: Refractive index changes optical path of ray of light forming fringe pattern.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 - B `beta = (lambda D)/(d) but lambda= (1)/(mu)` |
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673. |
Light travels a certain distance in water in `3 mu` s . How much time it would take for light for travel the same distance in air ? Refractive index of water = 4/3. |
Answer» Correct Answer - `2.2 mu` s |
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674. |
The equation of two light waves are `y_(1)=6cosomegat,y_(2)=8cos (omegat+phi)`.The ratio of maximum to minimum intensities produced by the supersposition of these waves will be-A. `49:1`B. `1:49`C. `1:7`D. `7:1` |
Answer» Answer: `a_1=6` units `a_2=8` units `(I_(max))/(I_(min))=[(a_(1))/(a_(2))+1]^(2)/[(a_(1))/(a_(2))-1]^(2)=[(6)/(8)+1]^(2)/[(6)/(8)+1]^(2)rArr (I_(max))/(I_(min))=(49)/(1)` |
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675. |
Which of the following is the path difference for destructive interference ?A. `n(lambda+1)`B. `(2n+1)(lambda)/(2)`C. `n lambda`D. `(n+1)(lambda)/(2)` |
Answer» Correct Answer - B For destructive interference, the path difference should be an odd multiple of `(lambda)/(2)`. |
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676. |
Ratio waves originating from sources `S_(1)` and `S_(2)` having zero phase difference and common wavelength `lambda` will show completely destructive interference at a point `P` is `S_(1)P-S_(2)P` isA. `5lambda`B. `3lambda//4`C. `2lambda`D. `11lambda//2` |
Answer» Answer: For destructive interference: Path difference =`S_(1)P-S_(2)P=(2n-1)lambda//2` `n=1,S_(1)P-S_(2)P=(2xx1-1)lambda//2=lambda//2` `n=2,S_(1)P-S_(2)P=(2xx2-1)lambda//2=3lambda//2` `n=3,S_(1)P-S_(2)P=(2xx3-1)lambda//2=5lambda//2` `n=4,S_(1)P-S_(2)P=(2xx4-1)lambda//2=7lambda//2` `n=5,S_(1)P-S_(2)P=(2xx5-1)lambda//2=9lambda//2` `n=6,S_(1)P-S_(2)P=(2xx6-1)lambda//2=11lambda//2` So,destructive pattern is possible only for path difference=`11lambda//2`. |
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677. |
What is the path difference of destructive interferenceA. `nlambda`B. `n(lambda+1)`C. `((n+1)lambda)/2`D. `((2n+a)lambda)/2` |
Answer» Correct Answer - D For destructive interference, `cos phi=min=-1` i.e., `phi=pmpi,pm3pi, pm 5pi` or `phi=pm(2n+1)pi` with `n=0, 1, 2, 3...` or `Deltax=pm((2n+1)lambda)/2` |
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678. |
If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima isA. `25:16`B. `5:3`C. `16:1`D. `25:9` |
Answer» Correct Answer - C In interference, `I_(max)=(sqrt(I_(1))+sqrt(I_(2)))^(2)` `"while " I_(max)=(sqrtI_(1)-sqrt(I_(2)))^(2)` `"So "I_(max)/I_(min)=(sqrt(I_(1))+sqrtI_(2))^(2)/((sqrt(I_(1))-sqrt(I_(2)))^(2))` `=(A_(1)+A_(2))^(2)/((A_(1)-A_(2))^(2))` :. Given `A_(1)/A_(2)=3/5` `:. I_(max)/I_(min)=((3+5)^(2))/((3-5)^(2))=16/1` |
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679. |
Two sources of sound of the same frequency produce sound intensities `I` and `4I` at a point `P` when used individually. If they are used together such that the sounds from them reach `P` with a phase differenceof `2pi//3`, the intensity at `P` will beA. 2lB. 3lC. 4lD. 5l |
Answer» Correct Answer - B `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi` `=I+4I+2sqrt(Ixx4I)xx"cos"(2pi)/3` `=5I+4Ixx(-1/2)=5I-2I=3I` |
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680. |
To make the central fringe at the center`O` , mica sheet of refractive index `1.5` is introduced Choose the corect statement. A. the thickness of sheet is `2(sqrt(2)-1)d` infront of `S_(1)`B. the thickness of sheet is `(sqrt(2)-1)d` infront of `S_(2)`C. the thickness of sheet is `2sqrt(2)d` infront of `S_(1)`D. the thickness of sheet is `(2sqrt(2)-1)d` infront `S_(1)`. |
Answer» Correct Answer - A Here the shift produced by mica sheet of thickness `t` is `(mut-t)=t(mu-1)`. It should be equal to extra path traveled by the ray `SS_(2)O` i.e., `Delta=SS_(2)-d=sqrt(2)d-d` `implies(t)/(2)=d(sqrt(2)-1)impliest=2d(sqrt(2)-1)` infront of `S_(1)` |
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