680 + Interview Questions in WAVE-OPTICS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Two identical coherent sources of wavelength `lambda` are placed at `(100 lambda, 0)` and `(-50 lambda, 0)` along x-axis. The number of maxima and minima detected are ,respectively [include origin and `(5 lambda, 0)]`A. 51 and 50B. 101 and 100C. 49 and 50D. 50 and 49
Answer» Correct Answer - b If the detector move by `0.5 lambda` from origin, it observers 2nd maxima. So, after every `0.5 lambda`, one maxima will be observed.

Discussion

152.	plane waves refracted for air to water using Huygen.s principal `a,b,c,d,e` ae length on the diagram.The refractive index of water wrt air is the ratio. A. `a//e`B. `b//e`C. `b//d`D. `d//b`
Answer» Correct Answer - A

Discussion

153.	Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in `d` remains constant.B. the absolute error in `d` increses.C. the fractional error in `d` remains constant.D. the fractional error in `d` decreases.
Answer» Correct Answer - D `2d sintheta=lambda` `d=(lambda)/(2 sin theta)` differntiate `del(d)=(lambda)/(2) del (cosectheta)` `del(d)=(lambda)/(2)(-cos esthetacottheta)deltheta` `del(d)=(-lambdacostheta)/(2sin^(2)theta)deltheta` as `theta= "increases", (lambdacostheta)/(2sin^(2)theta) , "decreases"` Alternate solution `d=(lambda)/(2sintheta)` `ln d=ln lambda-ln2-ln sintheta` `(Delta(d))/(d)=0-0-(1)/(sin theta)xxcostheta(Deltatheta)` Fraction error `\|+(d)\|=\|cotthetaDeltatheta\|` Absoulute error `Deltad=(dcottheta)Deltatheta` `(d)/(2sintheta)xx(costheta)/(sintheta)` `Deltad=(costheta)/(sin^(2)theta)`

Discussion

154.	Figure shows plane waves refracted for air to water. What is the refractive index of water with respect to air ? A. a/eB. b/eC. b/dD. d/b
Answer» Correct Answer - C

Discussion

155.	The figure shows `XY` separating two transparent media, medium-`1` and medium-`2.` The lenes ab and `cd` represent wavefronts of a light wave travelling in medium-`1` and incident on `XY.` The lines `rf` and `gh` represent wavefronts of the light wave in medium-`2` after refraction. The phases of the light wave at `c,d,e` and `f` are `phi_(c) ,phi_(d),phi_(e)` and `phi_(f)` respectively. It is given that `phi_(c)=phi_(f)` :A. `phi_(c)` cannot be equal to `phi_(d)`B. `phi_(d)` cannot be equal to `phi_(e)`C. `(phi_(d)-phi_(f))` is equal to `(phi_(c)-phi_(e))`D. `(phi_(d)-phi_(c))` is equal to `(phi_(f)-phi_(e))`
Answer» Correct Answer - C `c` and `d` are at same wavefronts So, `phi_(c)=phi_(d)` and `phi_(e)=phi_(f)` `therefore phi_(d)-phi_(f)=phi_(c)-phi_(f)` `D` is incorrect `phi_(d)-phi_(c)=0` `phi_(f)-phi_(e)=0` So both should be equal.

Discussion

156.	Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in `d` remains constant.B. the absolute error in `d` increses.C. the fractional error in `d` remains constant.D. the fractional error in `d` decreases.
Answer» Correct Answer - D `2d sintheta=lambda` `d=(lambda)/(2 sin theta)` differntiate `del(d)=(lambda)/(2) del (cosectheta)` `del(d)=(lambda)/(2)(-cos esthetacottheta)deltheta` `del(d)=(-lambdacostheta)/(2sin^(2)theta)deltheta` as `theta= "increases", (lambdacostheta)/(2sin^(2)theta) , "decreases"` Alternate solution `d=(lambda)/(2sintheta)` `ln d=ln lambda-ln2-ln sintheta` `(Delta(d))/(d)=0-0-(1)/(sin theta)xxcostheta(Deltatheta)` Fraction error `\|+(d)\|=\|cotthetaDeltatheta\|` Absoulute error `Deltad=(dcottheta)Deltatheta` `(d)/(2sintheta)xx(costheta)/(sintheta)` `Deltad=(costheta)/(sin^(2)theta)`

Discussion

157.	When light is refracted into a mediumA. its wavelength and frequncy both increaseB. its wavelength increase and frequncy remains unchangedC. its wavelength decrease and frequncy remains unchangedD. its wavelength and frequncy both decrease
Answer» Correct Answer - C

Discussion

158.	Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton. Speed of light isA. the same in medium-`1` and medium-`2`B. larger in medium-`1` than in medium-`2`C. larger in medium-`2` than in medium-`1`D. different ab `b` and `d`
Answer» Correct Answer - B Ray is bending towards normal. So, medium `1` should be rarer and medium `2` should be denser. So, `V_(1)gtV_(1)`.

Discussion

159.	Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton. Speed of light isA. the same in medium -1 and medium-2B. the larger in medium -1than the medium-2C. the larger in medium -2than the medium-1D. different at`b`and `d`
Answer» Correct Answer - B

Discussion

160.	Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton. The phase of the ligth wave at c, d, e, and f are `phi_(c)`, phi_(d), `phi_(e)` and `phi_(f)`, respectively. It is given that `phi_(c ) != phi_(f)`. ThenA. `phi_(c)`cannot be equal to`phi_(d)`B. `phi_(d)` can be equal to`phi_(c)`C. `(phi_(d)-phi_(f))` is equal to`(phi_(c)-phi_(e))`D. `(phi_(d)-phi_(c))` is not equal to`(phi_(f)-phi_(e))`
Answer» Correct Answer - C

Discussion

161.	Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton. Speed of light isA. the same in medium 1 and medium 2B. larger I medium 1 than in medium 2C. larger in medium 2 than in medium 1D. different at b and d.
Answer» Correct Answer - B

Discussion

162.	In YDSE shown in figure a parallel beam of light is incident on the slits from a medium of refractive index `n_(1)`. The wavelength of light in this medium is `lambda_(1)`. A transparent of thickness t and refractive index `n_(3)` is put in front of one slit. The medium between the screen and the plane of the slits is `n_(2)`. The phase difference between the light waves reaching point O (symmetrical, relative to the slits) is A. `(2pi)/(n_(1)lambda_(1))(n_(3)-n_(2))t`B. `(2pi)/(lambda_(1))(n_(3)-n_(2))t`C. `(2pin_(1))/(n_(2)lambda_(1))((n_(3))/(n_(2))-1)t`D. `(2pin_(1))/(lambda_(1))(n_(3)-n_(1))t`
Answer» Correct Answer - A

Discussion

163.	In a given YDSE setup if the wavelength of light beam incident ons lit plane is decreased then which of the following statement is/are correct :A. The intensity of bright fringes will decreaseB. Fringe pattern will shrinkC. Total number of bright fringes appearing on screen will increaseD. Only central fringe intensity will increase
Answer» Correct Answer - A

Discussion

164.	An unpolarized beam of light is incident on a group of three polarizing sheets which are arranged in such a way that plane of rotation of one make an angle of `60^(@)` with the adjacent one. The temperature of incident light transmitted by first polarizer will be :A. 6.25B. 12.5C. 50D. None of these
Answer» Correct Answer - C

Discussion

165.	Two identical sources each of intensity `I_(0)` have a separation `d = lambda // 8`, where `lambda` is the wavelength of the waves emitted by either source. The phase difference of the sources is `pi // 4` The intensity distribution `I(theta)` in the radiation field as a function of `theta` Which specifies the direction from the sources to the distant observation point P is given byA. `I(theta) = I_(0)cos^(2)theta`B. `I(theta)=(I)_(0)/(4)cos^(2)((pi theta)/(8))`C. `I(theta)=4I_(0)cos^(2)((pi)/(8)(sin theta+1))`D. `I(theta)= I_(0) sin^(2)theta`
Answer» Correct Answer - C `Delta x = d sin theta = (lambda)/(8) sin theta` `Delta phi = (2pi)/(lambda)xx(lambda)/(8) sin theta + Delta phi_(0)` (`Delta phi_(0)` is initial phase difference between the slits) `Delta phi = (pi)/(4) sin theta + (pi)/(4)` `Delta phi = (pi)/(4) (1+sin theta)` Resultant intensity `I = 4I_(0) cos^(2) ((Delta phi)/(2))` `I = 4I_(0) cos^(2) {(pi)/(8) (1+sin theta)}`

Discussion

166.	The path difference between two interfering waves at a point on the screen is `lambda // 6`, The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)A. 0.853B. 8.53C. 0.75D. 7.5
Answer» Correct Answer - c At path difference `(lambda)/(6)`, phase difference is `(pi)/(3)`. `I = I_(0) + I_(0) + 2I_(0) cos ((pi)/(3)) = 3 I_(0)` So, the required ratio is `(3 I_(0))/(4 I_(0)) = 0.75`.

Discussion

167.	Waves emitted by two identical sources produces intensity of K unit at a point on screen where path difference between these waves is `lamda`. Calculate the intensity at that point on screen at which path difference is `(lamda)/(4)`
Answer» `phi_(1)=(2pidelta)/(lamda)implies(2pi)/(lamda)xxlamda=2pi` and `phi_(2)=(2pi)/(lamda)xx(lamda)/(4)=(pi)/(2)I_(1)=I_(0)+I_(0)+2sqrt(I_(0))sqrt(I_(0))cos2pi=4I_(0)` and `I_(2)=I_(0)+I_(0)+2sqrt(I_(0))sqrt((I_(0))(cos2pi)/(2))=2I_(0)` `therefore(I_(1))/(I_(2))=(4I_(0))/(2I_(0))=2impliesI_(2)=(I_(1))/(2)=(K)/(2)` unit `[becauseI_(1)=K"unit"`]

Discussion

168.	The ratio of intensities of two waves are given by `4:1`. The ratio of the amplitudes of the two waves isA. `2:1`B. `1:2`C. `4:1`D. `1:4`
Answer» Correct Answer - A We know `I alpha A^(2)` `rArr (I_(1))/(I_(2))=(A_(1)^(2))/(A_(2)^(2))rArr sqrt(4/1)=(A_(1))/(A_(2)) rArr A_(1):A_(2) =2:1`

Discussion

169.	A slit 5 cm wide when irradiated by waves of wavelength 10 mm results in the angular spread of the central maxima on either side of incident light by about :A. `1 // 2` radianB. `1//4` radianC. 3 radianD. `1//5` radian
Answer» Correct Answer - A

Discussion

170.	The Brewster angle for the glass air interface is `54.74^(@)` if a ray of light going from air to glass strikes at an angle of incidence `45^(@)` then the angle of refraction isA. `60^(@)`B. `30^(@)`C. `25^(@)`D. `54.74^(@)`
Answer» Correct Answer - B `mu=tani_(p)=tan54.74^(@)=sqrt(2)` `because sqrt(2)=(sin 45^(@))/(sin r)rArr sin r=1/2rArr r=30^(@)`

Discussion

171.	If the polarizing angle of a piece of glass for green light is `54.74^(@)`, then the angle of minimum deviation for an equilateral prism made of same glass is : `["Given" : tan 54.74^(@)=1.414]`A. `45^(@)`B. `54.74^(@)`C. `60^(@)`D. `30^(@)`
Answer» Correct Answer - B

Discussion

172.	In YDSE, bichromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness isA. `4mm`B. `5.6mm`C. `14mm`D. `28mm`
Answer» Correct Answer - D Let nth minima of 400 nm coincides with mth minima of 560 nm, then `(2n - 1) ((400)/(2)) = (2m - 1) ((560)/(2))` or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10) =` … i.e., `4^(th)` minima of 400 nm coincides with `3^(rd)` minima of 560 nm. Location of this minima is, `Y_(1) = ((2xx4-1)(100)(400xx10^(-6)))/(2xx0.4) = 14 nm` Next `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm, then `(2n-1) ((400)/(2)) = (2m - 1) ((560)/(2))` or `(2n-1)/(2m-1) = (7)/(5) = (14)/(10)= ` ... i.e., `11^(th)` minima of 400 nm coincides with `8^(th)` minima of 560 nm. Location of this minima is, `+Y_(2)` = `((2xx11-1)(100)(400xx10^(-6)))/(2xx0.1) = 42mm` Required distance `= Y_(2)-Y_(1) = 28 mm` Hence, the correct optical is (D).

Discussion

173.	In the set up shown in figure, the two slits `S_1` and `S_2` are not equidistant from the slit S. The central fringe at O is then A. (a) Always brightB. (b) Always darkC. (c) Either dark or bright depending on the position of SD. (d) Neither dark nor bright
Answer» Correct Answer - C If path difference `Delta=(SS_1+S_1O)-(SS_2+S_2O)=nlambda`, `n=0, 1, 2, 3, …` the central fringe at O is a bright fringe and if the path difference `Delta=(n-1/2)lambda`, `n=1, 2, 3, …` the central bright fringe will be a dark fringe.

Discussion

174.	If `I_0` is the intensity of the principal maximum in the single slit diffraction pattern. Then what will be its intensity when the slit width is doubled?A. (a) `I_0`B. (b) `I_0/2`C. (c) `2I_0`D. (d) `4I_0`
Answer» Correct Answer - D If you divide the original slit into N strips and represents the light from each strip, when it reaches the screen, by a phasor, then at the central maximum in the diffraction pattern you add N phasors, all in the same direction and each with the same amplitude. The intensity is therefore `N^2`. If you double the slit width, you `2N` phasors, if they are each to have the amplitude of the each to have the amplitude of the phasors you used for the narrow slit. The intensity at the central maximum is proportional to `(2N)^2` and is, therefore, four times the intensity for the narrow slit.

Discussion

175.	If `I_0` is the intensity of the principal maximum in the single slit diffraction pattern. Then what will be its intensity when the slit width is doubled?A. `I_(0)`B. `I_(0)//2`C. `2I_(0)`D. `4I_(0)`
Answer» Correct Answer - 1

Discussion

176.	If `I_0` is the intensity of the principal maximum in the single slit diffraction pattern. Then what will be its intensity when the slit width is doubled?A. `2I_(0)`B. `4I_(0)`C. `(I_(0))/(2)`D.
Answer» Correct Answer - C `I=I_(0)((sintheta)/(theta))^(2)` and `theta=(pi)/(lamda)((ay)/(D))` for principal maximum `y=0,theta=0` Hence, intensity will remain same i.e., `I=I_(0)`

Discussion

177.	In YDSE, find the thickness of a glass slab `(mu=1.5)` which should be placed in front of the upper slit `S_1` so that the central maximum now lies at a point where 5th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is `5000 Å`.A. `5 xx 10^(-6)m`B. `3 xx 10^(-6)m`C. `10 xx 10^(-6) m`D. `5 xx 10^(-5) m`
Answer» Correct Answer - a According to the question, shift = 5 fringe widths `implies ((mu - 1)t D)/(d) = (5 lambda D)/(d)` `:.t = (5 lambda)/(mu - 1) = (25000)/(1.5 - 1) = 50,000 Å`

Discussion

178.	In YDSE, find the thickness of a glass slab `(mu=1.5)` which should be placed in front of the upper slit `S_1` so that the central maximum now lies at a point where 5th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is `5000 Å`.
Answer» Correct Answer - 50,000 `Å` According to the question, shift = 5 (fringe width) `:. ((mu -1)tD)/(d) = (5 lambda D)/(d)` `:. t = (5 lambda)/(mu - 1) = (25000)/(15.1) = 50,000 Å`

Discussion

179.	Plane of polarisation isA. The plane in which vibrations of the electric vector takes placeB. A plane perpendicular to the plane in which vibrations of the electric vector takes placeC. Is perpendicular to the plane of vibrationD. Horizontal plane
Answer» Correct Answer - A

Discussion

180.	Calculate the amount of sugar dissolved in `10^(-3)m^(3)` of water so as to produce ratation of plane of polarisation of `12^(@)`. Given the specific rotation of sugar solution is`0.01 rad m^(-1) kg^(-1) m^(3)` and length of polarimeter tube is 0.21m
Answer» Correct Answer - 0.01kg

Discussion

181.	A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?A. (a) Diffraction pattern is not observed on the screen in the case of electronsB. (b) The angular width of the central maximum of the diffraction pattern will increaseC. (c) The angular width of the central maximum will decreaseD. (d) The angular width of the central maximum will be unaffected
Answer» Correct Answer - C `c=vlambda` hence `vprop1/lambda`

Discussion

182.	A beam of light of wavelength `600nm` from a distance source falls on a single slit `1.00mm` side and the resulting diffraction pattern is observed on a screen `2m` away. The distance between the first dark frings on either side of the central bright fringe isA. (a) `1.2cm`B. (b) `1.2mm`C. (c) `2.4cm`D. (d) `2.4mm`
Answer» Correct Answer - D `(2Dlambda)/(d)=(2xx2xx600xx10^-9)/(1xx10^-3)m` `=24xx10^-4m=2.4mm`.

Discussion

183.	A beam of light of `lambda=600nm` from a distance source falls on a single slit `1mm` wide and the resulting diffraction pattern is observed on a screen `2m` away. The distance between first dark fringes on either side of the central bright fringe isA. (a) `1.2cm`B. (b) `1.2mm`C. (c) `2.4cm`D. (d) `2.4mm`
Answer» Correct Answer - D Given `D=2m`, `d=1mm=1xx10^-3m` `lambda=600nm=600xx10^-6m` Width of central bright fringe `(=2beta)` `=(2lambdaD)/(d)=(2xx600xx10^-6)/(1xx10^-3)m` `=2.4xx10^-3m=2.4mm`

Discussion

184.	A beam of light of wavelength `600nm` from a distant source falls on a single slit `1mm` wide and the resulting diffraction pattern is observed on a screen `2m` away. The distance between the first dark fringes on either side of the central bright fringe isA. 1.2 cmB. 1.2 cmC. 2.4 cmD. 2.4 mm
Answer» Correct Answer - B

Discussion

185.	A beam of light of wavelength `600nm` from a distant source falls on a single slit `1mm` wide and the resulting diffraction pattern is observed on a screen `2m` away. The distance between the first dark fringes on either side of the central bright fringe isA. (a) `1.2mm`B. (b) `1.2cm`C. (c) `2.4cm`D. (d) `2.4mm`
Answer» Correct Answer - D Distance between the first dark fringes on either side of central maxima=width of central maxima `=(2lambdaD)/(d)=(2xx600xx10^-9xx2)/(1xx10^-3)=2.4mm.`

Discussion

186.	Assertion: Radio waves can be polarised. Reason: Sound waves in air are longitudinal in nature.A. (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 - B Radio waves can be polarised becomes they are transverse in nature. Sound waves in air are longitudinal in nature

Discussion

187.	A flint glass and a crown glass are fitted on the two slits of a double slit apparatus. The thickness of the strips is `0.40 mm` and the separation between the slits is `0.12 cm`. The refractive index of fint glass and crown glass are `1.62` and `1.52` respectively for the light of wavelength `480 nm` which is used in the expweiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the geometrical centre will the nearest maximum be located ?
Answer» Correct Answer - (a) `beta=4.0xx10^(-4)m` (b) `(beta)/(3)and(2beta)/(3)`

Discussion

188.	The polaroids are placed in the path of unpolarized beam of intensity `I_(0)` such that no light is emitted fromthesecond polarid. If a third polaroid whose polarization axis makes an angle `theta` with the polarization axis of first polaroid, is placed between these4 polariods then the intensity of light emerging from the last polariod will beA. `((I_(0))/(8))sin^(2)2theta`B. `((I_(0))/(4))sin^(2)2theta`C. `((I_(0))/(2))cos^(2)theta`D. `I_(0)cos^(4)theta`
Answer» Correct Answer - A `I = (I_(0))/(2)cos^(2)theta`

Discussion

189.	Consider the diffraction pattern for a small pinhole. As the size of the hole is increasedA. the size decreaseB. the intensity increaseC. the size increaseD. the intensity decrease
Answer» Correct Answer - A::B (a,b) (a) When a decrease w increases. So, size decreases. (b) Now, light energy is distributed over a small are and intensity `prop(1)/("area")` as area is decreasing so intensity increases.

Discussion

190.	Which one of the following phenomena is not explained by Huygens construction of wavefront?A. RefractionB. RefrlectionC. DiffractionD. Formation of spectrum
Answer» Correct Answer - D

Discussion

191.	Which one of the following phenomena is not explained by Huygens construction of wavefront?A. (a) RefractionB. (b) ReflectionC. (c) DiffractionD. (d) Origin of spectra
Answer» Correct Answer - D Origin of spectra is not explained by Huygens theory.

Discussion

192.	For the same objective, the ratio of least separation between two points to be distiguised by a microscope for light of `5000 Å` and electrons acclerated through 100V used as illuminating substance is __________ (neraly)A. `2`B. `4xx10^(-2)`C. `2xx10^(-6)`D. `2xx10^(-4)`
Answer» Correct Answer - D Resolving power `= (1)/(Delta x) = (2sin alpha)/(1.22lambda)` `implies Delta x = (1.22lambda)/(2sin alpha)` For `lambda = 5000 Å Delta x_(1) = (1.22xx5000xx10^(-10))/(25sin alpha)` …(1) de-Brogle wavelength `lambda = (12.27)/(sqrt(v)) = (12.27)/(sqrt(100))` `lambda = 1.227xx10^(-10)` `Delta x_(2) = (1.22xx1.227xx10^(-10))/(2sin alpha)` `(Delta x_(1))/(Delta x_(2)) = (1.227xx10^(-10))/(5000xx10^(-10)) = 2xx10^(-4)`

Discussion

193.	Consider sunlight incident on a pinhole of width `10^(3)Å`. The image of the pinhole seen on a screen shall beA. be a fine sharp slit white in colour ast the center.B. a bright slit white at the center diffusing to zero intesities at the edges.C. a bright slit white at the center diffusing to regions of different colours.D. only be diffused slit white in colour
Answer» Correct Answer - A Consider for differaction is width of the slit `d le lambda`

Discussion

194.	Consider the diffraction patern for a small pinhole. As the size of the hole is increasedA. the size decreases.B. the intensity increasesC. the size increasesD. the intensity decreases.
Answer» Correct Answer - A::B `beta = 2lambda (D)/(d)` is the width of primary maximum in difraction pattern. Intensity `I = I_(0) ((sin alpha)/(alpha))^(2)` where `alpha = (pi d)/(lambda) sin theta`

Discussion

195.	Consider sunlight incident on a pinhole of width `10^(3)Å`. The image of the pinhole seen on a screen shall beA. a sharp white ringB. different from a geometrical imageC. a diffused central spot, white in colourD. diffused coloured region around a sharp central white spot
Answer» Correct Answer - B::D (b,d) Given, width of pinhole `=10^(3)Å=1000Å` We know that wavelength of sunlight ranges from `4000Å " to " 8000Å` Clearly, wavelength ` lambda lt ` width of the slit. Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed on the screen will be different from the geometrical image.

Discussion

196.	Consider sunlight incident on a pinhole of width `10^(3)Å`. The image of the pinhole seen on a screen shall beA. a sharp white ring.B. different from a geometrical image.C. a diffused central spot, white in colour.D. diffused coloured region around a sharp central white spot.
Answer» Correct Answer - B::D Condition for differaction is width of the slit `d le lambda`

Discussion

197.	Consider sunlight incident on a slit of width `10^(4) Å` . The image seen through the slit shallA. be a fine sharp slit white is colour at the centreB. a bright slit white at the centre diffusing to zero intensities at the edgesC. a bright slit white at the centre diffusing to regions of different coloursD. only be a diffused slit white in colour
Answer» Correct Answer - A (a) Given, width of the slit `=10^(4)` Å `=10^(4)xx10^(-10)m=10^(-6) m=1 mu m` Wavelength of ( visible ) sunlight varies from 4000Å to 8000Å. As the width of slit is comparable to that of wavelength, hence diffraction occcurs with maxima at centre. So, at the central all colours appear i.e., mixing of colours form white patch at the centre.

Discussion

198.	In a double slit experiment the angular width of a fringe is found to be `0.2^(@)` on a screen placed I m away. The wavelength of light used in 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be `4//3`.A. `0.2^(@)`B. `0.27^(@)`C. `0.3^(@)`D. `0.15^(@)`
Answer» Correct Answer - D

Discussion

199.	In a double slit experiment the angular width of a fringe is found to be `0.2^(@)` on a screen placed I m away. The wavelength of light used in 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be `4//3`.
Answer» Distance of the screen from the slits ,D=1m wavelength of light used ,`lamda_(1) =600 nm` Angular width of the fringe in ` air thera_(1) = 0.2 ^(@)` Angular of the fringe in water `= theta_(2)` Refractive index of water `mu =(4)/(3)` Refactive index is related to angular width as : `mu =(thata_(1))/(thera_(2))` `theta _(2) =(3)/(4)theta_(1)` `=(3)/(4) xx0.2 =0.15 ` therefore , the angluar width of the fringe in water will reduce to `0.15 ^(@)`

Discussion

200.	In double-slit experiment using light of wavelength 600 nm, the angular width of a fring formed on a distant screen is `0.1^(@)`. What is the spacing between the two slits ?
Answer» Here, `lambda=600nm=6xx10^(-7)m, theta=0.1^(@)=(0.1^@))/(180^(@))xxpirad,d=?` Since `theta=(lambda)/(d) Rightarrow d=(lambda)/(theta)=(6xx10^(-7)xx180^(@))/(0.1^(@)xxpi)=3.43xx10^(-4)m`

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