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301.

The width of central maximum is the…………….between……….on either side of…………… .

Answer» distance , first secondary minimum , centre of the pattern.
302.

Angular width of central maximum in diffraction at a single slit is……………. .

Answer» Correct Answer - `2 lambda//a`
303.

The angle of minimum deviation for prism of angle `pi//3 is pi//6`. Calculate the velocity of light in the material of the prism if the velocity of light in vacuum is `3 xx 10^8 ms^-1`.

Answer» Correct Answer - `2.121 xx 10^(8)ms^(-1)`
`mu = (sin(A + delta_(m))//2)/(sin A//2) = (sin (60^(@) + 30^(@))//2)/(sin 60^(@)//2)`
`= (sin 45^(@))/(sin 30^(@)) = (0.7070)/(0.5000) = 1.414`
`v = (c )/(mu) = (3 xx 10^(8))/(1.414) = 2.121 xx 10^(8) ms^(-1)`
304.

A transparent sphere of radius R and refractive index μ is kept in air. At what distance from the surface of the sphere should a point object be placed so as to form a real image at the same distance from the sphere?(a) R/μ(b) μR(c) R / μ - 1 (d) R / μ + 1

Answer»

Correct Answer is: (c) R / μ - 1 

From symmetry, the rays must pass through the sphere, parallel to the optic axis.

Using μ2/v - μ1/u = μ2 - μ1 / r

μ/  - 1/u = μ - 1 / R

u = - R / μ - 1.

305.

As per the recent understanding, light consists of (A) rays (B) waves (C) corpuscles (D) photons obeying the rules of waves

Answer»

Correct answer is (D) photons obeying the rules of waves

306.

Consider the optically denser lenses P, Q, R, and S drawn below. According to the Cartesian sign convention which of these have a positive focal length?(A) Only P (B) Only P and Q (C) Only P and R (D) Only Q and S

Answer»

Correct answer is (B) Only P and Q

307.

what type of wavefront will emerge from (i) a point source and (ii) distant light source ?A. converging sphericalB. Diverging sphericalC. planeD. cyclindrical

Answer» Correct Answer - C
308.

When a wave undergoes reflection at a denser medium, what happens to its phase ?A. `0^(@)`B. `45^(@)`C. `90^(@)`D. `180^(@)`

Answer» Correct Answer - D
309.

When a wave undergoes reflection at a denser medium, what happens to its phase ?

Answer» Reflection of waves at a denser medium causes a phase change of `pi` or `180^(@)`.
310.

If a wave undergoes refraction, what happens to its phase ?

Answer» No phase change occurs during refraction.
311.

In the specturm of white light through a prism, violet colour is seen at the bottom. Why ?

Answer» Violet colour is seen at the bottom because its wavelength is minimum and its deviation on passing through the rpism is maximum.
312.

What is the purpose of adding "blue" to clothes ?

Answer» On washing, clothes get a yellowish tint. Blue and yellow are comlimentary colours giving us a sensation of white.
313.

An isotropic point source emits light with wavelength `lambda = 589 nm`. The radiation power of the source is `P = 10W`. Find: (a) the mean density of the flow of photons at a distance `r = 2.0 m` from the osurce, (b) the distance between the source and the point at which the mean concentration of photons is equal to `n = 100 cm^(-3)`.

Answer» (a) The means density of the flow of photons at a disatnce `r` is
`lt j gt = (P)/(4pir^(2))//(2picancelh c)/(lambda) =(P lambda)/(8pi^(2)cancelh cr^(2))m^(-2)s^(-2)`
`= (10xx589xx10^(-6))/(8pi^(2)xx1.054xx10^(-34)xx10^(8)xx4)m^(-2)s^(-1)`
`= (10xx.589)/(8pi^(2)xx1.054xx12)xx10^(16)cm^(-2)s^(-1)`
`= 5.9xx10^(13)cm^(-2)s^(-1)`
(b) If `n(r)` is the mean concentration (number per unit volume) of photons at a distacne `r` form the source, then since all photons are moving outwards with a velocity `c`, there is an outward flux of `cn` which is balanced by the flux form the source. In steady state, the two are equal and so
`n(r) = (lt j gt)/(c ) = (P lambda)/(8pi^(2)cancelh c^(2)r^(2)) = n`
or `r = (1)/(2pic) sqrt((P lambda)/(2cancelh n))`
`= (1)/(6pixx10^(8)) sqrt((10xx589xx10^(-6))/(2xx1.054xx10^(-34)xx10^(2)xx10^(6)))`
`= (10^(2))/(6pi) sqrt((5.89)/(2.108)) = 8.87` metre
314.

Fig. shows the plot of the function `y(x)` representing a fraction of the total power of thermal radiation falling within the spectral interval from `0` to `x`. Here `x = lambda//lambda_(m)(lambda_(m)` is the wavelength corresponding to the maximum of spectral radiation density). Using this plot, find: (a) the wavelength which divides the radiation spectrum into two equal (in terms of energy) parts at the temperature `3700K`, (b) the fraction of the total radiation power falling within the visible range of the spectrum `(0.40-0.76 mu m)` at the temperature `5000K`, (c) how manu times the power radiated at wavelengths exceeding `0.76mu m` will increase if the temperature rises form `3000` to `5000K`.

Answer» (a) From the curve of the function `y(x)` we see that `y = 0.5` when `x = 1.41`
Thus `lambda = 1.41 xx (0.29)/(3700)cm = 1.105mum`
(b) At `5000K`
`lambda = (0.29)/(5)xx10^(-6)m = 0.58mu m`
So the visible range `(0.40` to `0.70)mu m` corresponds to a range `(0.69` to `1.31)` of `x`. From the curve
`y(0.69) = 0.07`
`y(1.31) = 0.44`
so the fraction is `0.37`
(c ) The value of `x` corresponding to `0.76` are
`x_(1) = 0.76//(0.29)/(0.3) = 0.786` at `3000K`
`x_(2) = 0.76//(0.29)/(0.5) = 1.31` at `5000K`
The requisite fraction is then
`((P_(2))/(P_(1))) = underset("ratio of t otal power")underset(uarr)(((T_(2))/(T_(1)))^(4))xxunderset("ratio of the fra ction of required wavel eng ths i n the radiated power")underset(uarr)((1-y_(2))/(1-y_(1)))`
`= ((5)/(3))^(4) xx(1-0.44)/(1-0.12) = 4.91`
315.

A beam of linearly polarized light is changed into a circularly polarized light by passing it through a slice of crystal 0.003 cm thick. Calculate the difference in the refractive index of the two rays in the crystal assuming this to be the minimum thickness that will produce the effect and that the wavelength is

Answer»

We know that, when we use quantum wave plate then a linear polarized beam change circular polarized beam.

t1/4 \(\cfrac14\)\(\cfrac{\lambda}{(H_0-H_e)}\)

where, H0 = refractive index of ordinary wave

H= refractive index of extra ordinary wave.

H0 - He\(\cfrac{\lambda}{4t_{t/4}}\)

H0 - He\(\cfrac{6\times10^{-5}}{4\times0.003}\)

H0 - He\(\cfrac{6\times10^{-2}}{12}\)

H0 - He = 0.5 x 10-2

316.

Find the wavelength of `X`-ray radiation if the maximum kinetic energy of compton electrons is `T_(max) = 0.19 MeV`.

Answer» We see form the previous problem that the electron gains the maximum `K.E.` when the photon is scattered backwards `theta = 180^(@)`. Then
`omega_(0) = omega_(0) = (mc^(2)//cancel h)/(sqrt(1+(2mc^(2))/(T_(max)) )-1)`
Hence `lambda_(0) = (2pic)/(omega_(0)) = (2picancel h)/(mc) [sqrt(1+(2mc^(2))/(T_(max)))-1]`
Substituting the values we get `= lambda_(0) = 3.695 pm`.
317.

Derive the formula for angle of deviation for thin prisms.ORShow that in a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

Answer»

For thin prisms (refracting angle < sin 10°).

sin θ ≈ θ

∴ Refractive index, n = \(\frac{sin\,i}{sin\,r_i}\) ≈ \(\frac{i}{r_1}\)

Also n = \(\frac{sin\,e}{sin\,r_2}\) ≈ \(\frac{e}{r_2}\)

∴ i ≈ n r1 and e ≈ nr2

ii. Substituting this in, i + e = A + δ, 

we get,

i + e = n (r1 + r2) = nA = A + δ

∴ S = A(n – 1)

A and n are constant for a given prism. 

Thus, for a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

318.

A source emits electromagnetic waves of wavelength `3 m`. One beam reaches the observer directly and other after reflection from a watersurface, travelling `1.5 m` extra distance and with intensity reduced to `1//4` as compared to intensity due to the direct beam alone. The resultant intensity will be :A. `(1//4) rad`B. `(3//4) rad`C. `(5//4) rad`D. `(9//4) rad`

Answer» Correct Answer - D
As `Delta x = 1.5 m = (3)/(2)m = (lambda)/(2) :. Delta phi = pi` ,br. Reflection from water surface introduces a phase diff. of `pi`. Therefore, total phase diff. `= pi + pi = 2pi`.
`:.` Intensity obtained is maximum.
Now `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 2pi`
`= I + (I)/(4) + 2sqrt(I xx I//4) = (9 I)/(4)`
319.

The total magnification produced by a compound microscope is `20`. The magnification produced by the eye piece is `5`. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be `14 cm`. If least distance of distinct vision is `20 cm`, calculate the focal length of objective and eye piece.

Answer» From `m = m_(0) xx m_(e)`,
`m_(0) = (m)/(m_(e)) = (20)/(5) = 4`
Now, `m_(0) = (v_(0))/(u_(0)) = (L)/(f_(0)) = 4`,
`f_(0) = (L)/(4) = (14)/(4) = 3.5 cm`
Also, `m_(e) = 1 + (d)/(f_(e)) = 5`
`(d)/(f_(e)) = 5-1 = 4, f_(e) = (d)/(4) = (20)/(4) = 5 cm`
320.

A compound microscope has an objective of focal length `1 cm` and an eye piece of focal length `2.5 cm` away frm the object has to be placed at a distance of `1.2 cm` away from the objective for normal adjustment. Find the angular mignification and length of the microscope tube.

Answer» Here, `f_(0) = 1 cm, f_(e) = 2.5 cm`.
`u_(0) = - 1.2 cm, m = ? L = ?`
As `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))`
`:. (1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(1) - (1)/(1.2) = (0.2)/(1.2)`
`v_(0) = 1.2//0.2 = 6 cm`
As `m = (v_(0))/(|u_(0)|)(1 + (d)/(f_(e)))`
`:. m = (6)/(1.2)(1 + (25)/(2.5)) = 55`
`L = v_(0) + f_(e) = 6 + 2.5 = 8.5 cm`
321.

A person uses `+ 1.5 D` glasses to have normal vision from `25 cm` onwards. He uses a `20 D` lens as a simple microscope to see an object. Calculate the maximum magnifying power, if he uses the microscope (a) together with his glasses (b) without the glasses.

Answer» Here, `P_(1) = 1.5 D : d = 25 cm`,
`P_(2) = 20 D`
(a) with glasses on
`M = 1 + (D)/(f_(2)) = 1 + (25)/(100//20) = 6`
(b) without glasses, Focal length of glasses,
`f_(1) = (100)/(P_(1)) = (100)/(1.5) = (200)/(3)cm`.
`u = - 25 cm, v = ?`
As `(1)/(v) - (1)/(u) = (1)/(f_(1))`
`:. (1)/(v) = (1)/(f_(1)) + (1)/(u) = (3)/(200) - (1)/(25) = (3 - 8)/(200) = (-5)/(200)`
`v = - 40 cm`.
`M = 1 + (v)/(f_(2)) = 1 + (40)/(100//20) = 9`
322.

In YDSE, light of wavelength `5000 Å` is used. The third bright band on the screen is formed at a distance of `1 cm` from the central bright band. If the screen is at a distance of `1.5 m` from the centre of narrow slits, calculate the separation between the slits.

Answer» Correct Answer - `2.25 xx 10^(-4) m`
Here,`lambda = 5000 Å = 5 xx 10^(-7)m, n =3`
`x_(3) = 1 cm = 10^(-2)m, D = 1.5m, d = ?`
`x_(3) = n (lambda D)/(d)`
`10^(-2) = 3 xx 510^(-7) xx (1.5)/(d) = (22.5 xx 10^(2))/(d)`
`d = (22.5 xx 10^(-7))/(10^(-2)) = 22.5 xx 10^(-5)m`
`= 2.25 xx 10^(-4)m`
323.

A certain oscillation results from the addition of coherent oscillations of the same direction results from the addition of coherent where `k` is the number of the oscillation `(k = 1, 2, ………., N), varphi` is the phase difference between the `kth` and `~(k -1)th` oscillations. Find the amplitude of the resultant oscillation.

Answer» We use the method of complex amplitudes. Then the ampulitudes are
`A_(1) = a, A_(2) = ae^(i varphi), ..A_(N) = ae^(i(N - 1)varphi)`
and the resultant complex amplitude is
`A = A_(1) + A_(2) +………+A_(N) = a (1 -e^(i varphi) + e^(2i varphi)+.......+e^(i(N - 1)varphi))`
`= a(1 - e^(iN varphi))/(1 -e^(i varphi))`
The correspnding orbinary amplitude is
`|A| = a |(1 - e^(iN varphi))/(1 - e^(i varphi))| = a[(1 - e^(iNvarphi))/(1 - e^(i varphi)) xx (1 - e^(-iN varphi))/(1 -e^(-i varphi))]^(1//2)`
`= a[(2- 2 cos N varphi)/(2 - 2 cos varphi)]^(1//2) = a (sin((Nvarphi)/2))/(sin((varphi)/(2)))`.
324.

The near vision of an average person is `25 cm`. To view an object with an angular magnification of `10`, what should be the power of the microscope ?

Answer» Here, `d = 25 cm, m = 10, P = ?`
Angular magnification, `m = (d)/(f) :. f = (d)/(m) = (25)/(10) = 2.5 cm P = (100)/(f) = (100)/(2.5) = 40 D`
325.

The number of lenses in terrestrial telescope is (a) 2 (b) 4 (c) 3 (d) 6

Answer»

(c) 3 

A terrestrial telescope consists of three lenses: 

objective, erecting lens and eyepiece.

326.

What is the use of an erecting lens in a terrestrial telescope?

Answer»

A terrestrial telescope has an additional erecting lens to make the final image erect.

327.

What is the use of collimator?

Answer»

The collimator is an arrangement to produce a parallel beam of light.

328.

The collimator is an arrangement to produce a parallel beam of light.

Answer»

The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.

329.

What is myopia? What is its remedy?

Answer»

Myopia (or) short sightedness:

It is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.

Remedy (correction):

A myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

330.

For a normal eye, the least distance of distinct vision is……..and far point is………… .

Answer» `25 cm` , at infinity
331.

What is hypermetropia? What is its remedy?

Answer»

Hypermetropia (or) Long sightedness: 

It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly. 

Remedy (correction): 

A hypermetropic eye is corrected by using a convex lens of suitable focal length.

332.

Presbyopia is also called………. .

Answer» Correct Answer - old sight
333.

What is presbyopia?

Answer»

This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).

334.

What is astigmatism?

Answer»

Astigmatism is the defect arising due to different curvatures along different planes in the eye lens. Astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.

335.

What is meant by astigmatism ?

Answer»

In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the . refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses (Torrid lenses).

336.

Assertion : Raman scattering the scattering of monochromatic light by atoms and molecule of a liquid.Reason: The wavelength of Raman lines is same.(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 Assertion is false but Reason is true.

Answer»

(b) If both Assertion and reason are true but Reason is not the correct explanation of Assertion.

337.

What is elastic scattering ?

Answer»

If the energy of the incident beam of light and the scattered beam of light are same, then it is called as 'elastic scattering'.

338.

What is scattering of light?

Answer»

When sunlight enters the Earth’s atmosphere, the atoms and molecules of different gases present in the atmosphere refract the light in all possible directions. This is called as ‘Scattering of light’.

339.

What is inelastic scattering ?

Answer»

If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as 'inelastic scattering'.

340.

State the types of scattering.

Answer»

1. Elastic scattering

2. Inelastic scattering

341.

Sonu has observed some lines in solar spectrum are absorbed by the elements present in the atmosphere. What are the lines?

Answer»

The lines are Fraunhofer lines.

342.

Assertion: Raman spectrum of a liquids contains lines whose frequencies are not equal to that of incident radiation. Reason: If a photon strikes an atom in a liquid that is in existed state photon losses energy. (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 Assertion is false but Reason is true.

Answer»

(b) If both Assertion and Reason are true but reason is not the correct explanation of Assertion.

343.

What are Raman lines?

Answer»

The spectral lines which are having frequencies other than the incident ray frequency are called 'Raman lines'.

344.

Assertion: Light waves travel in straight lines. Reason: Rectilinear propagation of light confirm the above mentioned properly. (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 Assertion is false but Reason is true.

Answer»

(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion

345.

What are stokes lines and Antistokes  lines ?

Answer»

The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequency higher than the incident frequency are called antistokes lines.

346.

How are different types of scattering formed? Mention the types of scattering.

Answer»

The nature and size of the scatterer results in different types of scattering. 

They are 

1. Rayleigh scattering 

2. Mie scattering 

3. Tyndall scattering 

4. Raman scattering

347.

Mention the advantages of telescope.

Answer»

1. Elaborate view of the Galaxies, Planets, stars and other heavenly bodies is possible. 

2. Camera can be attached for taking photograph for the celestial objects. 

3. Telescope can be viewed even with the low intensity of light.

348.

Assertion : A single lens produces a coloured image of an object illuminated by white light.Reason: The refractive index of material of lens is different for different wavelength of light. (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 Assertion is false but Reason is true.

Answer»

(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

349.

How does refractive index (μ) of a material vary with respect to wavelength (λ). (A and B are constants).(a) μ = A + \(\frac{B}{λ^2}\) (b) μ = A + Bλ2(c) μ = A +\(\frac{B}{λ}\) (d) μ = A + Bλ

Answer»

(a) μ = A + \(\frac{B}{λ^2}\) 

According to cauchy’s relation, (a) μ = A + \(\frac{B}{λ^2}\) 

350.

What are primary focus and secondary focus of convex lens?

Answer»

Primary focus: The primary focus F1 is defined as a point where an object should be placed to give parallel emergent rays to the principal axis. 

Secondary focus: The secondary focus F2 is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.