804 + Interview Questions in OPTICS IN GENERAL AWARENESS Page 10 InterviewSolution

451.	Does nature of the image depend upon size of the mirror?
Answer» No, nature of the image is independent of size of the mirror.

Discussion

452.	Out of electric field vector, `vec(E)` and magnetic field vector, vec(B)` in an electromagnetic wave, which is more effective and why ?A. `vec(E)`B. `vec(B)`C. Both `vec(E)` and `vec(B)`D. neither `vec(E)` nor `vec(B)`
Answer» Correct Answer - A

Discussion

453.	What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?
Answer» When a ray of light is incident on a plane mirror at an angle θ, the reflected ray gets deviated by an angle of (π – 2θ).

Discussion

454.	Calculate the refractive index of the material of an equilaterial prism for which angle of minimum deviation is `60^@`.
Answer» Correct Answer - `sqrt(3)` Here, `A = 60^(@), delta_(m) = 60^(@), mu = ?` `mu = (sin (A + delta_(m))//2)/(sin A//2) = (sin (60^(@) + 60^(@))//2)/(sin 60^(@)//2)` `= (sin 60^(@))/(sin 30^(@)) = (sqrt(3)//2)/(1//2) = sqrt(3)`

Discussion

455.	A plane wave front falls on a convex lens. The emergent wave front isA. planeB. diverging sphericalC. converging sphericalD. none of these
Answer» Correct Answer - C

Discussion

456.	The sun is visible to us before actual………….and after………. . This is because of ………………. .
Answer» sunrise , set , atmoshperic refraction.

Discussion

457.	Find the relation between the group velocity `u` and phase velocity `upsilon` for the following disperison laws: (a) `upsilon prop 1//sqrt(lambda_(1))`, (b) `upsilon prop k`, (c ) `upsilon prop 1// omega^(2)`, Here `lambda, k`, and `omega` are the wavelength, wave number, and angular frequancy.
Answer» (a) `v = a//sqrt(lambda), a =` constant Then `u = v- lambda (dv)/(d lambda)` `= (a)/(sqrt(lambda)) - lambda (-(1)/(2)a lambda^(-3//2)) = (3)/(2).(a)/(sqrt(lambda)) = (3)/(2)v`. (b) `v = bk = omegak, b =` constant so `omega = bk^(2)` and `u = (d omega)/(dk) = 2bk = 2v`. (c) `v = (c)/(omega^(2)), c =` constant `= (omega)/(k)`. so `omega^(3) = ck` or `omega = c^(1//3) k^(1//3)` Thus `u = (d omega)/(dk) = c^(1//3) (1)/(3)k^(-2//3) = (1)/(3)(omega)/(k) = (1)/(2)v`

Discussion

458.	Reflection is the phenomenon of………without…….. .
Answer» change in the path of light , any change in medium

Discussion

459.	Total internal reflection of light is the phenomenon of………..into……..from……….. .
Answer» reflection of light , denser medium , interface of denser and rarer medium.

Discussion

460.	For toat internal reflection, light must travel……………to………….. .
Answer» from densermedium , rarer medium.

Discussion

461.	What is the ratio of speed of `IR` rays and `UV` rays in vacuum ?
Answer» `1 : 1` i.e. speed in vaccume is the same for both `IR and UV` rays.

Discussion

462.	In a certain medium the relationship between the group and phase velocities of an electromagnetic wave has the form `uv = c^(2)`, where `c` is the velocity of light in vaccume. Find the dependence of permittivity of that medium on wave frequency, `epsilon(omega)`.
Answer» We have `uv = (omega)/(k)(d omega)/(dk) = c^(2)` Intergrating we find `omega^(2) = A+c^(2) k^(2), A` is a constant. so `k = sqrt(omega^(2) - A)/(c )` and `v = (omega)/(k) = (c )/(sqrt(1- (A)/(omega^(2))))` writing this as `c//sqrt(epsilon(omega))` we get `epsilon(omega) =1-(A)/(omega^(2))` `(A` can be `+ve` or negative)

Discussion

463.	Can we obtain the image formed by a convex mirror on a screen ? If not, why ?
Answer» No, because the image formed by a convex mirror is virtual.

Discussion

464.	Two plane mirrors parallel to each other and an object O placed between them. Then the distance of the first three images from the mirror M2 will be (in cm) – A. `5, 10, 15`B. `5, 15, 30`C. `5, 25, 35`D. `5, 15, 25`
Answer» Correct Answer - C

Discussion

465.	Out of electric field vector, `vec(E)` and magnetic field vector, vec(B)` in an electromagnetic wave, which is more effective and why ?
Answer» `vec(E)` is more effective than `vec(B)`. This is because when a charge `q` moving with a velocity `v` encounters an e.m. wave, `(F_(e))/(F_(m)) = (q E)/(q v B) = (E)/(v B) = ( c)/(v) [ because (E)/(B) = c]` As `c gt gt v, :. F_(e) gt gt F_(m)`

Discussion

466.	An object is placed between two parallel plane mirrors. Why do the distance images get fainter and fainter ?
Answer» Images are fromed due to multiple reflections. At each reflection, a part of light energy is absorbed. Therefore, distance images get fainter.

Discussion

467.	Image formed in a concave mirror may be…………depending upon……… .
Answer» real or virtual , the position of the object.

Discussion

468.	A beam of `X`-rays impinges on a three-dimensional rectangular array whose periods are `a, b,` and `c`. The direction of the incident beam coincides with the direction along which the array period is equal to `a`. Find the directions to the diffraction maxima and the wavelength at which these maxima will be observed.
Answer» Suppose `alpha, beta`, and `gamma` are the angles between the direction to the diffraction maximum and the directions of the array along the periods `a, b,` and `c` respectively (call them `x, y, & z` axes). Then the value of these angles can be found from the following familiar condiations `a(1 - cos alpha) = k_(1) lambda` `b cos beta = k_(2) lambda` and `c cos gamma = k_(3) lambda` where `k_(1), k_(2), k_(3)` are whole numbers `(+, -,` or `0`) (These formules are, in effect, Laue equations, see any text book on modern physics). Squaring and adding we get on using `cos^(2) alpha + cos^(2) beta + cos^(2) gamma = 1` `2-2 cos alpha [((k_(1))/(a))^(2) + ((k_(2))/(b))^(2) + ((k_(3))/(c))^(2)]lambda^(2) = (2k_(1)lambda)/(a)` Thus `lambda = (2k_(1)a)/([[k_(1)//a)^(2) +(k+_(2)//a)^(2) +(k_(3)//a)^(2)]]`. Knowing `a, b, c` and the interger `k_(1), k_(2), k_(3)` we can find `alpha, beta, gamma` as well as `lambda`.

Discussion

469.	The linear magnification of a convex mirror is always…………because image formed in such a mirror is always………. .
Answer» positive , virtual and erect.

Discussion

470.	Optical fibers are based on the phenomenon ofA. reflectionB. refractionC. dispersionD. total internal relflection
Answer» Correct Answer - D

Discussion

471.	What are different types of dispersions of light? Why do they occur?
Answer» i. There are two types of dispersions : a. Angular dispersion b. Lateral dispersion ii. The refractive index of material depends on the frequency of incident light. Hence, for different colours, refractive index of material is different. iii. For an obliquely incident ray, the angles of refraction are different for each colour and they separate as they travel along different directions resulting into angular dispersion. iv. When a polychromatic beam of light is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours separated out. v. In this case, these colours are parallel to each other and are also parallel to their initial direction resulting into lateral dispersion

471.

What are different types of dispersions of light? Why do they occur?

Answer»

i. There are two types of dispersions :

a. Angular dispersion

b. Lateral dispersion

ii. The refractive index of material depends on the frequency of incident light. Hence, for different colours, refractive index of material is different.

iii. For an obliquely incident ray, the angles of refraction are different for each colour and they separate as they travel along different directions resulting into angular dispersion.

iv. When a polychromatic beam of light is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours separated out.

v. In this case, these colours are parallel to each other and are also parallel to their initial direction resulting into lateral dispersion

Discussion

472.	Find the maximum kinetic enegry of photoelectrons liberated form the surface of lithium by electromagnetic radiation whose electric component varies with time as `E = a(1+cos omegat) cos omega_(0)t`, where `a` is a constant, `omega = 6.0.10^(14)s^(-1)` and `omega_(0) = 360.10^(15)s^(-1)`.
Answer» We write `E =a (1+cos omegat) cos omega_(0)t` `= acos omega_(0)t +(a)/(2)[cos(omega_(0)-omega)t+cos(omega_(0)+omega)t]` It is abvious that light has there frequencies and the maximum `K.E.` of photon electrons ejected is `h(omega +omega_(0))-A_(Li)` where `A_(Li) = 2.39eV`. Substituting we get `0.37ev`.

Discussion

473.	Find the kinetic enegry of electrons emitting light in a medium with refractive index `n = 1.50` at an angle `theta = 30^(@)` to their propagation direction.
Answer» From `cos theta = (v)/(V)` we get `V = V sec theta` so `(V)/(c ) = (v)/(c )sec theta = (sec theta)/(n) = (sec 30^(@))/(1.5) = (2//sqrt(3))/(3//2) = (4)/(3sqrt(3))` Thus for electrons `T_(e) = 0.511 [(1)/(sqrt(1-(16)/(27)))-1] = 0.511 [sqrt((27)/(11))-1] = 0.289 MeV` Generally `T = mc^(2)[(1)/(sqrt(1-(1)/(n^(2)cos^(2)theta)))-1]`

Discussion

474.	The temperature of one of the two heated black bodies is `T_(1) = 2500K`. Find the temperature of the other body if the wavelength corresponding to its maximum emissive capacity exceeds by `Delta lambda = 0.50mu m` the wavelength corresponding to the maximum emissive capacity of the first black body.
Answer» For the first black body `(lambda_(min))_(1) = (b)/(T_(1))` Then `(lambda_(min))_(2) = (b)/(T_(1)) + Delta lambda = (b)/(T_(2))` Hence `T_(2) = (b)/((b)/(T_(1))+Delta lambda) = (bT_(1))/(b+T_(1)Delta lambda) = 1.747kK`

Discussion

475.	An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm. The final image is formed at infinity. The focal lengths fo of the objective and fe of the eyepiece are (a) 45 cm and -9 cm respectively (b) 50 cm and 10 cm respectively (c) 7.2 cm and 5 cm respectively (d) 30 cm and 6 cm respectively
Answer» Correct Answer is: (d) 30 cm and 6 cm respectively Magnification = 5 = f_o/f_e. Tube length = 36 = f_o + f_e.

475.

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm. The final image is formed at infinity. The focal lengths fo of the objective and fe of the eyepiece are (a) 45 cm and -9 cm respectively (b) 50 cm and 10 cm respectively (c) 7.2 cm and 5 cm respectively (d) 30 cm and 6 cm respectively

Answer»

Correct Answer is: (d) 30 cm and 6 cm respectively

Magnification = 5 = f_o/f_e.

Tube length = 36 = f_o + f_e.

Discussion

476.	Rainbow is seen only for a definite angle range with respect to the ground. Justify.
Answer» i. For clear visibility of rainbow, a beam must have enough intensity. ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity. iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

476.

Rainbow is seen only for a definite angle range with respect to the ground. Justify.

Answer»

i. For clear visibility of rainbow, a beam must have enough intensity.

ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity.

iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

Discussion

477.	What is polarising angle ?
Answer» It is the angle of incidence of unpolarised light on a transparent refracting medium, corresponding to which, reflected light is plane polarised completely.

Discussion

478.	Two concave lenses each of focal length `30 cm` are placed in contact. What is focal of the compound lens ?
Answer» `-15 cm`, from `1//f = 1//f_(1) + 1//f_(2)`

Discussion

479.	Two thin lenses of power `+ 6 D and - 2 D` are in contact. What is the focal length of the combination ?
Answer» `P = P_(1) + P_(2) = 6 - 2 = 4 D` `F = (100)/(P) = (100)/(4) = 25 cm`

Discussion

480.	What is the basis of an optical fibre ?
Answer» It is based on the phenomenon of total internal reflection of light.

Discussion

481.	One dioptre is the power of a lens of facal lengthA. `1 cm`B. `1 m`C. `-1 cm`D. `- 1m`
Answer» Correct Answer - B

Discussion

482.	what is the deviation produced in ray passing through optical centre of the lens ?
Answer» Correct Answer - Zero degree.

Discussion

483.	A lens forms a virtual, erect and diminished image whatever be the position of the object. Which type of lens this ?
Answer» The lens must be a concave lens.

Discussion

484.	What is a polaroid ?
Answer» A polaroid is a device that produces an intense beam of plane polarised light.

Discussion

485.	A transparent sphere of radius `R` and refractie index `mu` is kept in air. At what distance from the surface of the sphere shold a point object be placed so as to form a rea image at the same distance from the sphere?A. `R//mu`B. `muR`C. `R/(mu-1)`D. `R/(mu+1)`
Answer» Correct Answer - C

Discussion

486.	Why can we not get diffraction pattern from a wide slit illuminated by monochromatic light ?
Answer» When slit is wide (i.e. `a gt gt lambda`), bending of light becomes so small that it cannot be detected upto a certain distance of screen from the slit. Hence, practically, no detectable diffraction occurs.

Discussion

487.	A `12 m` tall tree is to be photographed with a pin hole camera. It is situated `15 m` away from the pin hole. How far should the screen be placed from the pin hole to obtain a `12 cm` tall image of the tree ?
Answer» Correct Answer - `15 cm` Here, `h_(1) = 12 m, u = 15 m` `v = ? h_(2) = 12 cm = 0.2 m` As `(h_(2))/(h_(1)) = (v)/(u)` `:. V = (h_(2))/(h_(1)) xx u = (0.12)/(12) xx 15 = 0.15 m = 15 cm`

Discussion

488.	What is Brewster angle for air to glass transtion ? (`mu` of glass is `1.5`)
Answer» Here, `i_(p) = ?, mu = 1.5`. As `tan i_(p) = mu = 1.5 :. i_(p) = tan^(-1)(1.5)` `i_(p) = 56.3^(@)`

Discussion

489.	There are two thin symmetrical lenses : one is converging , with refractive index `n_(1) = 1.70`, and the other is diverging with refractive index `n_(2) = 1.51`. Both lenses have the same curavature close together and submerged into whater. What is the focal length of this system in water ?
Answer» Focal length of the converging lens, when it is submerged in water of `R.I. n_(0)` (say): `(1)/(f_(1)) = ((n_(1))/(n_(0)) - 1) ((1)/(R ) - (1)/(R )), (2(n_(1) - n_(0)))/(n_(0)R)` (1) Similarly, the focal length of divierging lens in water. `(1)/(f_(2)) = ((n_(1))/(n_(0)) - 1) ((1)/(-R ) - (1)/(R )), (-2(n_(1) - n_(0)))/(n_(0)R)` (2) Now, when they are put together in the water, the focal length of the system, `(1)/(f) = (1)/(f_(1)) + (1)/(f_(2))` `= (2(n_(1) - n_(2)))/(n_(0)R) - (2(n_(2) - n_(0)))/(n_(0)R) = (2(n_(1) - n_(2)))/(n_(0)R)` or, `f = (-n_(0)R)/(2(n_(1) - n_(2))) = 35 cm`

Discussion

490.	Distance between the sources is one meter. An observer observes minimum intensity at some point outside the line joining the sources, then the wavelength of the wave should be –A. 1 mB. 2mC. 0.5 mD. 0.25 m
Answer» Correct Answer - B

Discussion

491.	Mark the correct options :A. If the far point is `1m` away from the eye, diverging lens should be used,B. If the near point is `1 m` away from the eye, divergent lens should be used,C. If the far point goes ahead, the power of the divergent lens should be reduced,D. If the near point goes ahead, the power of the convergent lens should be reduced.
Answer» Correct Answer - A::C A diverging lens has to be used when far point is `1 m` away. Further, if far point shifts ahead, focal length of diverging lens should be increased or its power should be reduced.

Discussion

492.	A narrow beam of white light slab having parallel faces.A. The light inside the slab is white,B. The light inside the slab is spit into different colours,C. The emergent beam is white,D. The light never splits in different colours.
Answer» Correct Answer - B::C On passing through a slab having parallel faces, light inside the slab is split into different colours, but the emerging beam is white.

Discussion

493.	A ray of white light passes through a rectangular glass slab, entering and emerging at parallel faces. The angle of incidence, measured from the normal to the glass surface, is large. (a) White light will emerge from the slab. (b) The light emerging from the slab will have a number of parallel, coloured rays. (c) The emergent rays will not form a spectrum on a screen. (d) Colours will be seen if the emergent rays enter the eye directly.
Answer» Correct Answer is: (a, b, & )

Discussion

494.	A ray of white light passes through a rectantular glass slab, entering and emerging at parallel faces. The angle of incidence, measured from the normal to the glaass surface, is large.A. White light will emerge from the slab.B. The light emerging from the slab will have a number of parallel, coloured raysC. The emergent rays wil not form a spectrum on a screen.D. Colours will be seen if the emergent rays enter the eye directly.
Answer» Correct Answer - A::B::C

Discussion

495.	Why do some people use bifocal lenses ?
Answer» Some people suffer from both, myopia and hypermetropia. They require bifocal lenses.

Discussion

496.	A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?
Answer» Given: n_g = 1.52, n_w = 1.33 To find: Critical angle (i_c) formula: sin i_c = \(\frac{n_2}{n_1}=\frac{n_w}{n_g}\) Calculation: From formula, i_c = sin^-1 \((\frac{1.33}{1.52})\) = sin^-1 (0.875) = 61 °2′

496.

A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?

Answer»

Given: n_g = 1.52, n_w = 1.33

To find: Critical angle (i_c)

formula: sin i_c = \(\frac{n_2}{n_1}=\frac{n_w}{n_g}\)

Calculation:

From formula,

i_c = sin^-1 \((\frac{1.33}{1.52})\) = sin^-1 (0.875) = 61 °2′

Discussion

497.	A Kerr cell is positioned between two crossed Nicol prisms so that the direction of electric field `E` in the capacitor forms an angle of `45^(@)` with the principle directions of the prisms. The capacitor has the length `l = 10 cm` and is falled up with nitrobenzence. Light of wavelength `lambda = 0.50 mu m` passes through the system. Taking into account that in this case the kerr constant is equal to `B = 2.2.10^(-10)cm//V^(2)`, find: (a) the minimum strength of electric field `E` in the capacitor at which the intensity of light that passes through this syetm is independent of rotation of the rear prism, (b) how many times per second light will be interrupted when a sinusoidal voltage of frequecny `v = 10 MHz` and strength amplitude `E_(m) = 50kV//cm` is applied to the capacitor. Note. The Kerr constant is the coefficient `B` in the equation `n_(e) = n_(0) = Blambda E^(2)`.
Answer» In passing through the Kerr cell the two perpendicualr components of the electirc field will acquire a phase difference. When its phase difference quals `90^(@)` the emergent light will be circularly polarized because the two perpendicualr components `O & E` have the same magnitude since it is given that the direction of electric field `E` in the capacitor forms an angle of `45^(@)` with the principle directions of then icols. In this case the intensity of light that emerges from this system will be independent of the rotation of the analyser prism. Now the phase difference introduced is given by `delta = (2pi)/(lambda) (n_(e) - n_(0)) l` In the present case `delta = (pi)/(2)` (for minimum electric field) `n_(e) - n_(0) = (lambda)/(4l)` Now `n_(e) - n_(0) = B lambda E^(2)` so `E_(min) sqrt((1)/(4Bl)). = 10^(5)// sqrt(88) = 10.66 kV// cm`. (b) If the applied electric fiel is `E = E_(m) sin omegat, omega = 2pi v` then the Kerr cell introduces a time varying phase difference `delta = 2pi B\|E_(m)^(2) sin^(2) omegat` `= 2pi xx 2.2 xx 10^(-10) xx 10 xx (50 xx 10^(3))^(2) sin^(2)omegat` `= 11pi sin^(2) omega t` In one half-cycle (i.e. in time `(pi)/(omega) = T//2 = (1)/(2v)`) this reaches the value `2kpi` when `sin^(2) omegat = 0, (2)/(11), (4)/(11), (6)/(11), (8)/(11), (10)/(11)` `(2)/(11), (4)/(11), (6)/(11), (8)/(11), (10)/(11)` i.e. `11` times. On each be interrupted when the Kerr cell (placed between crossed Nicols) introduced a phase difference of `2 k pi` and in no other case.)

Discussion

498.	A square object is placed `15 cm` from a convex mirror of radius of curavture `90 cm`. Calculate the position of the image and its areal magnification.
Answer» Correct Answer - `11.25 cm` behind the mirror : `9//16` Areal magnification `= ("linear magnification")^(2)`

Discussion

499.	Monochromatic plane-polarized light with angular frequency `omega` through a certain substance along a unifrom magnetic field `h`. Find the difference of refractive indices for right-hand and left-hand components of light beam with circular polarization if the verdet consatnt is equal to `V`.
Answer» From problem `189`, we know that `Deltan = (alpha lambda)/(pi)` where `alpha` is the rotation constant. Thus `Deltan = (2alpha)/(2pi//lambda) = (2alphac)/(omega)` On the other hand `alpha_(mag) = VH` Thus for the magntic rotation `Deltan = (2c VH)/(omega)`

Discussion

500.	An intially parallel cyclindrical beam travels in a medium of refractive index `mu (I) = mu_(0) + mu_(2) I`, where `mu_(0)` and `mu_(2)` are positive constants and `I` is intensity of light beam. The intensity of the beam is decreasing with increasing radius. Answer the following questions : The speed of light in the medium isA. the same everywhere in the beamB. directly proportional to the intensity `I`C. maximum on the axis of the beamD. ( d) minimum on the axis of the beam
Answer» Correct Answer - D The speed of light in the medium is minimum on the axis of the beam, because refractive index of medium ia maximum at the axis of beam.

Discussion

Explore topic-wise InterviewSolutions in .

Does nature of the image depend upon size of the mirror?

Out of electric field vector, `vec(E)` and magnetic field vector, vec(B)` in an electromagnetic wave, which is more effective and why ?A. `vec(E)`B. `vec(B)`C. Both `vec(E)` and `vec(B)`D. neither `vec(E)` nor `vec(B)`

What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?

Calculate the refractive index of the material of an equilaterial prism for which angle of minimum deviation is `60^@`.

A plane wave front falls on a convex lens. The emergent wave front isA. planeB. diverging sphericalC. converging sphericalD. none of these

The sun is visible to us before actual………….and after………. . This is because of ………………. .

Find the relation between the group velocity `u` and phase velocity `upsilon` for the following disperison laws: (a) `upsilon prop 1//sqrt(lambda_(1))`, (b) `upsilon prop k`, (c ) `upsilon prop 1// omega^(2)`, Here `lambda, k`, and `omega` are the wavelength, wave number, and angular frequancy.

Reflection is the phenomenon of………without…….. .

Total internal reflection of light is the phenomenon of………..into……..from……….. .

For toat internal reflection, light must travel……………to………….. .

What is the ratio of speed of `IR` rays and `UV` rays in vacuum ?

In a certain medium the relationship between the group and phase velocities of an electromagnetic wave has the form `uv = c^(2)`, where `c` is the velocity of light in vaccume. Find the dependence of permittivity of that medium on wave frequency, `epsilon(omega)`.

Can we obtain the image formed by a convex mirror on a screen ? If not, why ?

Two plane mirrors parallel to each other and an object O placed between them. Then the distance of the first three images from the mirror M2 will be (in cm) – A. `5, 10, 15`B. `5, 15, 30`C. `5, 25, 35`D. `5, 15, 25`

Out of electric field vector, `vec(E)` and magnetic field vector, vec(B)` in an electromagnetic wave, which is more effective and why ?

An object is placed between two parallel plane mirrors. Why do the distance images get fainter and fainter ?

Image formed in a concave mirror may be…………depending upon……… .

The linear magnification of a convex mirror is always…………because image formed in such a mirror is always………. .

Optical fibers are based on the phenomenon ofA. reflectionB. refractionC. dispersionD. total internal relflection

What are different types of dispersions of light? Why do they occur?

Find the maximum kinetic enegry of photoelectrons liberated form the surface of lithium by electromagnetic radiation whose electric component varies with time as `E = a(1+cos omegat) cos omega_(0)t`, where `a` is a constant, `omega = 6.0.10^(14)s^(-1)` and `omega_(0) = 360.10^(15)s^(-1)`.

Find the kinetic enegry of electrons emitting light in a medium with refractive index `n = 1.50` at an angle `theta = 30^(@)` to their propagation direction.

Rainbow is seen only for a definite angle range with respect to the ground. Justify.

What is polarising angle ?

Two concave lenses each of focal length `30 cm` are placed in contact. What is focal of the compound lens ?

Two thin lenses of power `+ 6 D and - 2 D` are in contact. What is the focal length of the combination ?

What is the basis of an optical fibre ?

One dioptre is the power of a lens of facal lengthA. `1 cm`B. `1 m`C. `-1 cm`D. `- 1m`

what is the deviation produced in ray passing through optical centre of the lens ?

A lens forms a virtual, erect and diminished image whatever be the position of the object. Which type of lens this ?

What is a polaroid ?

A transparent sphere of radius `R` and refractie index `mu` is kept in air. At what distance from the surface of the sphere shold a point object be placed so as to form a rea image at the same distance from the sphere?A. `R//mu`B. `muR`C. `R/(mu-1)`D. `R/(mu+1)`

Why can we not get diffraction pattern from a wide slit illuminated by monochromatic light ?

A `12 m` tall tree is to be photographed with a pin hole camera. It is situated `15 m` away from the pin hole. How far should the screen be placed from the pin hole to obtain a `12 cm` tall image of the tree ?

What is Brewster angle for air to glass transtion ? (`mu` of glass is `1.5`)

There are two thin symmetrical lenses : one is converging , with refractive index `n_(1) = 1.70`, and the other is diverging with refractive index `n_(2) = 1.51`. Both lenses have the same curavature close together and submerged into whater. What is the focal length of this system in water ?

Distance between the sources is one meter. An observer observes minimum intensity at some point outside the line joining the sources, then the wavelength of the wave should be –A. 1 mB. 2mC. 0.5 mD. 0.25 m

A narrow beam of white light slab having parallel faces.A. The light inside the slab is white,B. The light inside the slab is spit into different colours,C. The emergent beam is white,D. The light never splits in different colours.

Why do some people use bifocal lenses ?

A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?

A square object is placed `15 cm` from a convex mirror of radius of curavture `90 cm`. Calculate the position of the image and its areal magnification.

Monochromatic plane-polarized light with angular frequency `omega` through a certain substance along a unifrom magnetic field `h`. Find the difference of refractive indices for right-hand and left-hand components of light beam with circular polarization if the verdet consatnt is equal to `V`.