804 + Interview Questions in OPTICS IN GENERAL AWARENESS Page 16 InterviewSolution

751.	A ray of light is incident at an angle of `60^(@)` on a horizontal plane mirror. Through what angle should the mirror be tilted to make the reflected ray horizontal ?
Answer» Correct Answer - `15^(@)` To make the reflected ray horizontal, it must be turned further through `90^(@) - 60^(@) = 30^(@)`. Therefore, the mirror must be turned through `30//2 = 15^(@)`

Discussion

752.	Out of speed, frequency and wavelength, name the parameter (s) which remain the same on reflection.A. speedB. frequencyC. wavelengthD. none of these
Answer» Correct Answer - B

Discussion

753.	A mirror is turned through `15^(@)`. Through what angle will the reflected ray turn ?
Answer» `30^(@)`, as the reflected ray turns through twice the angle thorugh which mirror is turned.

Discussion

754.	How would you combine two lenses of focal lengths `25 cm and 2.5 cm` to make a telescope ? What is the magnifying power of this telescope ?
Answer» Correct Answer - Lenses should be held at a distance of `27.5 cm` from eachother, `M = - 10`

Discussion

755.	What si the number of images of a point object held inbetween two plane mirrors inclined at an angle `theta^(@)` ?
Answer» Number of images , `n = ((360^(@))/(theta) - 1)`, if `(360)/(theta)` is even integer and `n = (360)/(theta)`, if `(360)/(theta)` is odd integer.

Discussion

756.	A reflecting type telescope has a concave reflector of radius of curvature `120 cm`. Calculate focal length of eye piece to secure a magnification of `20`.
Answer» Correct Answer - `3 cm`

Discussion

757.	Chromatic aberrations is caused due to(A) spherical shape of lens (B) spherical shape of mirrors (C) angle of deviation for violet light being more than that for red light. (D) refractive index for violet light being less than that for red light
Answer» (C) angle of deviation for violet light being more than that for red light.

Discussion

758.	(a) What is focal length of a convex mirror of radius of curvature `20 cm`? (b) What is radius of curvature of a mirror of focal length `- 50 cm` ?
Answer» (a) Here, `f = ? R = 20 cm` `f = (R )/(2) = (20)/(2) = 10 cm` (b) Here, `f = - 50 cm, R = ?` `R = 2f = 2 (- 50) = - 100 cm`

Discussion

759.	A point object is held between two plane mirror held at (i) `24^(@)` (ii) `30^(@)`. What is the number of images formed in the two cases ?
Answer» (i) When `theta = 24^(@)` `n = (360)/(theta) = (360)/(24) = 15` (ii) When `theta = 30^(@)` `n = (360)/(theta) -1 = (360)/(30) - 1 = 12 - 1 = 11`

Discussion

760.	Four light waves are represented by (i) `y=a_1sin omegat` (ii) `y=a_2 sin (omega t+varphi)` (iii) `y=a_1sin 2 Omega t` (iv) `y=a_2sin 2(omegat+varphi)` Interference fringes may be observed due to superposition ofA. (i) and (iii)B. (ii) and (iv)C. (i) and (ii)D. (iii) and (iv)
Answer» Correct Answer - A::B For interference, frequency of two waves must be same and phase difference must be constant, the amplitudes may be different.

Discussion

761.	When light propagates in vacuum there is an electric field and a magnetic field. These fieldsA. constant in timeB. mutually perpendicularC. having zero average valuesD. both perpendicular to the direction of propagation of light
Answer» Correct Answer - B::D Electric and magnetic fields in an electromagnetic (light) wave are mutually perpendicular and also perpendicular to the direction of propagation of light.

Discussion

762.	Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
Answer» Given: M.P = 12, v = D, f₀ = 90 cm, To find: Focal length of eye piece (f_e) Formula: M.P = \(\frac{f_0}{f_e}\) (1 + \(\frac{f_e}{D}\)) Calculation: From formula. 12 = \(\frac{90}{f_e}\) (1 + \(\frac{f_e}{25}\)) ∴ f_e = 10.71 cm

762.

Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?

Answer»

Given: M.P = 12, v = D, f₀ = 90 cm,

To find: Focal length of eye piece (f_e)

Formula: M.P = \(\frac{f_0}{f_e}\) (1 + \(\frac{f_e}{D}\))

Calculation:

From formula.

12 = \(\frac{90}{f_e}\) (1 + \(\frac{f_e}{25}\))

∴ f_e = 10.71 cm

Discussion

763.	You are given two converging lenses of focal lengths `1.25 cm and 5 cm` to design a compound microscope. If it is desired to have a mignification of `30`, find out separation between the objective and eye piece.
Answer» Here, `f_(0) = 1.25 cm, f_(e) = 5 cm,` `d = 25 cm, L = ? M = 30` When final image is formed at the near point of eye, the magnifying power of compound microscope is given by `M = (L)/(f_(0))(1 + (d)/(f_(e)))` `30 = (L)/(1.25)(1 + (25)/(5)) = (6 L)/(1.25)` `L = (30 xx 1.25)/(60 = 6.25 cm`

Discussion

764.	A compound microscope has lenses of focal length `10 mm and 30 mm`. An object placed at `1.2 cm` from the first lens is seen through the second lens at `0.25 m` from the eye lens. Calculate (i) magnifying power (ii) distance between the two lenses.
Answer» Correct Answer - `-46.7 ; 8.7 cm` Here, `f_(0) = 10 mm = 1 cm, f_(e) = 30 mm = 3 cm` `u_(0) = - 1.2 cm, v_(e) = - 0.25 m = - 25 cm` `M = ? L = ?` From `(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) = (1)/(6)` `v_(0) = 6 cm` From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))` `(1)/(u_(e)) - (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(3) = (-28)/(75)` `u_(e) = (- 75)/(28) = - 2.7 cm` `M = (v_(0))/(-u_(0))(1 + (d)/(f_(e))) = (6)/(-1.2)(1 + (25)/(3)) = -46.7` `L = v_(0) + \|u_(0)\| = 6 + 2.7 = 8.7 cm`

Discussion

765.

Complete the following table for a concave mirror?Position of objectPosition of imageLateral magnificationU = ∞v = fm = 0u > 2f…………….m < 1u = fV = ∞……………………………|v| > |u|m > 12f > u > f………………m > 1

Answer»

Position of object	Position of image	Lateral magnification
U = ∞	v = f	m = 0
u > 2f	2f > v > f	m < 1
u = f	V = ∞	m = ∞
u < f	\|v\| > \|u\|	m > 1
2f > u > f	v > 2f	m > 1

Discussion

766.	…………of spherical mirror is called……….. Of the mirror.
Answer» the diameter , aperture

Discussion

767.	With light falling normally on a transparent diffraction grating `10mm` wide, it was found that the components of the yellow line of sodium `(589.0` and `589.6nm`) are resolved beginning with the fifth order of the spectrum. Evaluate: (a) the period of this grating, (b) what must be the width of the grating with the same period for a double `lambda = 460.0nm` whose components differ by `0.13nm` to be resolved in the third order of the spectrum.
Answer» Here `R = (lambda)/(delta lambda) = (589.3)/(0.6) = kN = 5N` so `N = (589.3)/(3) = (10^(-2))/(d)` `d = (3 xx 10^(-2))/(589.3)m = 0509mm` (b) To resolve a doublet with `lambda = 460.0nm` and `delta lambda = 0.13nm` in the thrid order we must have `N = (R )/(3) = (460)/(3 xx 0.13) = 1179` This means that the grating is `Nd = 1179 xx 0.0509 = 60.03mm` wide `= 6cm` wide.

Discussion

768.	A transparent diffraction grating of a quartz spectrograph is `25mm` wide and has `250` lines per millimetre. The focal length of an objective in whose focal plane a photographic plate is located is equal to `80cm`. Light falls on the grating at right angles. The spectrum under investigating includes a doublet with components of wavelength `310.154` and `310.184nm`. Determine: (a) the disatnce on the photographic plate between the components of this doublet in the spectra of the first and the second order, (b) whether these components will be resolved in these orders of the spectrum.
Answer» From `d sin theta = k lambda` we get `delta theta = (k delta lambda)/(d cos theta)` On the other hand `x = fsin theta` so `delta x = f cos theta delta theta = (kf)/(d) delta lambda` For `f = 0.80m, delta lambda = 0.30nm` and `d = (1)/(250)mm` we get `delta x= {(6,mu,m,if,k,=,1),(12,mu,m,if,k,=,2):}` (b) Here `N = 25 xx 250 = 6250` and `(lambda)/(delta lambda) = (310.169)/(0.03) = 10339.. gtN` and so resolve we need `k = 2` For `k = 1` givens an `R.P.` of only `6250`.

Discussion

769.	Sound waves are not electromanetic waves asA. they can undergo interferenceB. they can undergo diffractionC. they cannot be polarizedD. they cannot pass through vacuum
Answer» Correct Answer - D

Discussion

770.	A cardsheet divided into squares each of size `1 mm^(2)` is being viewed at a distance of `9 cm` through a magnifying glass (a conerging lens of focal length `10 cm`) held close to the eye. (a) What is the magnification produced by the lenas ? How much is the area of each square to the virtual image ? (b) What is the angular magnification (magnifying power) of the lens ? ( c) Is the magnification in (a) equal to the magnifying power in (b) ? Explain
Answer» (a) Here, area of each (object) square `= 1 mm^(2), u = -9 cm, f = 10cm`. As `(1)/(v) - (1)/(u) = (1)/(f) :. (1)/(v) = (1)/(f) + (1)/(u) = (1)/(10) - (1)/(9) = (9 - 10)/(90) = -(1)/(90)` `v = -90 cm`. Magnification, `m = (v)/(\|u\|) = (90)/(9) = 10` `:.` Area of each square in virtual image `= (10)^(2) xx 1 = 100 sq. mm` (b) Magnifying power `= (d)/(u) = (25)/(9) = 2.8` ) c) No, magnification in (a) which is `(v//u)` cannot be equal to magnifying power in (b) which is `(d//u)` unless `v = d` i.e. image is located at the least distance of distinct vision.

Discussion

771.	A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in.A. a larger angle to be subtended by the object at the eye and hence viewed in greater detail.B. the formation of a virtual erect image.C. increase in the field of view.D. infinite magnification at the near point.
Answer» Correct Answer - A::B A magnifying glass is used as the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence viewed in greater detail. Also, it results in the formation of a virtual, erect image. Choices (a) and (b) are correct.

Discussion

772.	If ɛ0 and μ0 are the electric permittivity and magnetic permeability of free space respectively, and ɛ and μ are the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameter is(a) ɛμ/ɛ0μ0(b) (ɛμ/ɛ0μ0)1/2(c) (ɛ0μ0/ɛμ)(d) (ɛ0μ0/ɛμ)1/2
Answer» Correct Answer is: (b) (ɛμ/ɛ₀μ₀)^1/2 Velocity of light in vacuum = C = 1/√(μ₀ɛ₀) Velocity of light in the medium = v = 1/√μɛ. Refractive index = c/v = (μɛ/ɛ₀μ₀)^1/2

772.

If ɛ0 and μ0 are the electric permittivity and magnetic permeability of free space respectively, and ɛ and μ are the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameter is(a) ɛμ/ɛ0μ0(b) (ɛμ/ɛ0μ0)1/2(c) (ɛ0μ0/ɛμ)(d) (ɛ0μ0/ɛμ)1/2

Answer»

Correct Answer is: (b) (ɛμ/ɛ₀μ₀)^1/2

Velocity of light in vacuum = C = 1/√(μ₀ɛ₀)

Velocity of light in the medium = v = 1/√μɛ.

Refractive index = c/v = (μɛ/ɛ₀μ₀)^1/2

Discussion

773.	Fig. (a) and (b) show refraction of an incident ray in air at `60^@` with the normal to a glass-air and water-air interface respectively. Predict the angle of refraction of an incident ray in water at `45^@` with the normal to a water glass interface. Take `.^a mu_g = 1.32`. .
Answer» In Fig. `I = 60^(@), r = 35^(@)` `.^(a)mu_(g) = (sin i)/(sin r) = (sin 60^(@))/(sin 35^(@)) = (0.8660)/(0.5736) = 1.51` In Fig. `I = 60^(@), r = 41^(@)` `.^(a)mu_(w) = (sin i)/(sin r) = (sin 60^(@))/(sin 47^(@)) = (0.8660)/(0.7314) = 1.18` In Fig. `i = 45^(@), r = ?` `.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (sin i)/(sin r)` `(1.51)/(1.32) = (sin 45^(@))/(sin r) = (0.7071)/(sin r) :. sin r = (1.32 xx 0.7071)/(1.51) = 0.6181` `r = 38.2^(@)`

Discussion

774.	(a) What is the essential condition for Rayleigh scattering ? (b) In Rayleigh scattering, how is intensity of scattered light related to wavelength of light ?
Answer» When size of scatterer is much smaller than wavelength of light, then intensity of scattered light varies inversely as fourth power of wavelength of light, i.e., `I-(s) prop (1)/(lambda_(4))`.

Discussion

775.	From the data of four lenses given in `Q.4` which one will you select as objective of a compound microscope and which one as eye lens ? How can the magnifying power of such a microscope be increased ?
Answer» The objective of a compound microscope should have small focal length and small aperture. Therefore, lens `D` serves as the objective lens. For larger magnifying power, focal length of eye lens should be small (though greater than the focal length of objective lens). Therefore, lens `C` is selected as eye lens. Magnifying power, `m = (L)/(f_(0))(1 + (d)/(f_(e)))` `m = (L)/(5)(1 + (25)/(10)) = (7 L)/(10)` Clearly, `m` can be increased by increasing the length `L` of the microscope tube.

Discussion

776.	In a compound microscope, the distance between objective lens and eye lens is.A. fixedB. variableC. infiniteD. `1 metre`
Answer» Correct Answer - A

Discussion

777.	A compound microscope uses an objective lens of focal length `4 cm` and eye lens of focal length `10 cm`. An object is placed at `6 cm` from the objective lens. Calculate magnifying power of compound microscope if final image is formed at the near point. Also, calculate length of the tube of compound microscope.
Answer» Correct Answer - `7` ; `22 cm` Here, `f_(0) = 4 cm, f_(e) = 10 cm, u_(0) = - 6 cm` From `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))` `(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(4) - (1)/(6) = (1)/(12)` `v_(0) = 12 cm` `M = (v_(0))/(\|u_(0)\|)(1 + (d)/(f_(e))) = (12)/(6)(1 + (25)/(10)) = 7` Length of microscope `= L = v_(0) + f_(e)` `= 12 + 10 = 22 cm`

Discussion

778.	In a compound microscope, the objective lens and……..are………. .
Answer» eye lens , fixed distance apart.

Discussion

779.	Find the wavelength of the short-wave limit of an `X`-ray continuous spectrum if electrons approach the anticathode of the tube with velocity `v = 0.85 c`, where `c` is the velocity of light.
Answer» The wavelength of `X`-rays is the least when all the `K.E.` of the electrons approaching the anticathode is converted into the energy of `X-`ray. But the `K.E.` of electron is `T_(m) = mc^(2) [(1)/(sqrt(1-v^(2)//c^(2)))-1]` `(mc^(2) =` rest mass energy of electrons `= 0.511MeV`) Thus `(2pi cancelh c)/(lambda) = T_(m)` or `lambda = (2pi cancelh c)/(T_(m)) = (2pi cancelh)/(mc) [(1)/(sqrt(1-v^(2)//c^(2))-1)]^(-1)` `= (2picancelh)/(mc(gamma-1)), gamma = (1)/(sqrt(1-v^(2)//c^(2))) = 2.70 pm`

Discussion

780.	Convex lens is made of glass of refractive index `1.5` If the radius of curvature of each of the two surfaces is `20 cm` find the ratio of the powers of the lens, when placed in air to its power, when immersed in a liquid of refractive index `1.25`.
Answer» Correct Answer - `5//2` Here, `mu_(g) = 1.5, R_(1) = 20 cm = 0.2 m` `R_(2) = - 20 cm = - 0.2m` `P_(1) = (1)/(f_(1)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= (1.5 - 1)((1)/(0.2) + (1)/(0.2)) = 5 D` When the same lens is placed in liquid, `P_(2) = (1)/(f_(2)) = ((mu_(g))/(mu_(l)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= ((1.5)/(1.25) - 1)((1)/(0.2) + (1)/(0.2)) = (0.25)/(1.25) xx (2)/(0.2) = 2 D` `(P_(1))/(P_(2)) = (5)/(2)`

Discussion

781.	A double convex lens of glass of refractive index `1.6` has its both surfaces of equal radii of curvature of `30 cm `each. An object of height `5 cm` is placed at a distance of `12.5 cm` from the lens. Calculate the size of the image formed.
Answer» Correct Answer - `10 cm` Here, `mu = 1.6, R_(1) = 30 cm, R_(2) = - 30 cm` `h_(1) = 5 cm, u = - 12.5 cm, h_(2) = ?` `(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= (1.6 - 1)((1)/(30) + (1)/(30)) = (0.6)/(15)` `f = (15)/(0.6) = 25 cm` As `(1)/(v) - (1)/(u) = (1)/(f)` `:. (1)/(v) = (1)/(f) + (1)/(u) = (1)/(25) - (1)/(12.5) = (-1)/(25)` `v = - 25 cm` As `m = (h_(2))/(h_(1)) = (v)/(a) :. (h_(2))/(5) = (- 25)/(-12.5) = 2` `h = 10 cm`

Discussion

782.	In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is R. Determine the focal length of this system.
Answer» (i) For refraction at the first surface of the lens Here, `u = - oo, R = + R` Suppose the first surface converges the incident parallel beam at a distance `v_(1)`, assuming medium `mu_(2)` to be continuous. As `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )` `:. -(mu_(1))/(-oo) + (mu_(2))/(v_(1)) = (mu_(2)-mu_(1))/(R )` or `(mu_(2))/(v_(1)) = (mu_(2) - mu_(1))/(R )` ...(i) (ii) For refraction at second surface of the lens `-(mu_(2))/(u) + (mu_(3))/(v) = (mu_(3) - mu_(2))/(R )` ...(ii) The image formed by the first surface acts as virtual object for the second surafce and this second surface forms the image at the focus of the lens. `:. u = v_(1), v = + f and R = + R` From (ii), `-(mu_(2))/(v_(1)) + (mu_(3))/(+f) = (mu_(3) - mu_(2))/(R )` ...(iii) Adding (i) and (ii), we get `(mu_(3))/(f) = (mu_(2) - mu_(1))/(R ) + (mu_(3) - mu_(2))/(R ) = (mu_(3) - mu_(1))/(R )` `f = (mu_(3)R)/(mu_(3) - mu_(1))`

Discussion

783.	An astronomical telescope of magnifying power `7` consists of two thin lenses `40 cm` apart, in normal adjustment. Calculate the focal lengths of the lenses.
Answer» Here, `M = (f_(0))/(f_(e)) = 7 or f_(0) = 7 _(e)` In normal adjustment, distance between the lenses, `f_(0) + f_(e) = 40 cm` `:. 7 f_(e) + f_(e) = 40, f_(e) = (40)/(8) = 5 cm` `f_(0) = 7 f_(e) = 7 xx 5 = 35 cm`

Discussion

784.	A student has drawn the following courses of rays through a glass prism, Fig. Which one represents the position of minimum deviation ?A. B. C. D.
Answer» Correct Answer - D Choice (d) in Fig. shows the correct course of rays through prism in minimum deviation position. This is because in this position, the refracted ray is parallel to the base of the prism and incident ray and emergent ray are inclined equally to the two faces of the prism.

Discussion

785.	In Finding experimentally, the `mu` of a liquid using a convex lens and a plane mirror, parallax is removed whenA. object needle is at the slit focus of the lensB. object needle is at `2 F`C. object needle lies between `F` and `2 F`D. none of the above
Answer» Correct Answer - A Parallax is removed when object needle is at focus of the lens. On refraction through convex lens, rays are rendered parallel. They fall normally on the plane mirror, and retrace their path reflection. The image of the object is formed at only. That is why parallax is removed.

Discussion

786.	Use the mirror equation to show that an object placed between `f and 2f` of a concave mirror forms an image beyond `2f`.
Answer» Let `u = -(3 f)/(2)`, when object lies between f and 2f. `(1)/(v) + (1)/(u) = (1)/(f)` `(1)/(v) = (1)/(f) - (1)/(u) = (-1)/(f) + (2)/(3f) = (-3 + 2)/(3f) = -(1)/(3f)` For concave mirror, `f` is neg. `v = -3f` i.e., distacne of image from the concave mirror is `3f`, i.e., image formed is beyond `2f`.

Discussion

787.	Use the mirror equation to deduct that : (a) an object between `f` and `2f` of a concave mirror produces a real image beyond `2 f`. (b) a convax mirror always produces a virtual image independent of the location of the object. ( c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Answer» (a) The mirror equation is `(1)/(v) + (1)/(u) = (1)/(f)` or `(1)/(v) = (1)/(f) - (1)/(u)` For a concave mirror, `f` is negative i.e., `f lt 0`. As object is on the left, `u` is negative, i.e., `u lt 0` As object lies between `f` and `2f` of a concave mirror, `:. 2f lt u lt f` `(1)/(2f) gt (1)/(u) gt (1)/(f)` or `-(1)/(2f) lt -(1)/(u) lt -(1)/(f)` or `(1)/(f) - (1)/(2f) lt (1)/(f) - (1)/(u) lt 0` or `(1)/(2f) lt (1)/(v) lt 0` `:. (1)/(v)` is negative or `v` is negative. The image is real. Also `v gt 2f` i.e. the image lies beyond `2f`. (b) For a concave mirror, `f` is positive i.e., `f gt 0`. As object is on the left, `u` negative, i.e., `u lt 0`. As `(1)/(v) = (1)/(f) - (1)/(u)` `:. (1)/(v)` is positive or `v` is positive i.e., image is at the back of the mirror. Hence image is virtual, whatever be the value of `u`. ( c) For a concave mirror, `f gt 0` and `u lt 0` As `(1)/(v) = (1)/(f) - (1)/(u), therefore ((1)/(v)) gt ((1)/(f))` i.e. `v , f` `:.` image is located between the pole and the focus. As `v lt \|u\|`, the image is diminished. (d) For a concave mirror, `f lt 0` As object is placed between the pole and focus `:. f lt u lt 0 :. ((1)/(f) - (1)/(u)) gt 0` But `((1)/(f) - (1)/(u)) = (1)/(v) gt 0` or `v` is positive. Image is on the right, it must be virtual. Also, `(1)/(v) lt (1)/(\|u\|) i.e. v gt \|u\|` `:.` Image is enlarged.

Discussion

788.	A simple microscope is a combination of two lenses of power `+ 15 D` and `+ 5 D` in contact. Calculate magnifying power of microscope, if final image is formed at `25 cm` from the eye.
Answer» Here, `P = P_(1) + P_(2) = 15 + 5 = 20 D` Focal length of combination, `f = (100)/(P) = (100)/(20) = 5 cm` Magnifying power, `m = 1 + (d)/(f) = 1 + (25)/(5) = 6`

Discussion

789.	A right angled prism is to be made by selecting a proper material and the angles A and B `(B
Answer» (`a`) As already discussed `Agetheta_(c )`, `Bgetheta_(c )` Also `BleA` Case : `1 : B=A=pi//4rArrtheta_(c )lepi//4`…..(`1`) Case `2 : B lt A`. Hence, for reflection at both faces `B gt theta_(c )` But, `B lt pi//4rArrtheta_(c ) lt pi//4` (as `2B lt pi//2`).......(`2`) Combining Eqs. (`1`) and (`2`), we get `theta_(c ) le pi//4` `(1)/(n) le(1)/(sqrt(2))rArrn ge sqrt(2)` `n_(min)=sqrt(2)` (`b`) This pertains to case (`2`) Hence, `theta_(c ) lt 30^(@)` But `theta_(c )=sin^(-1)(3//5)` which is more than `30^(@)` and hence, it is not possible.

Discussion

790.	Two points `P` and `Q` lie on either side of an axis `XY` as shown. It is desired to produce an image of `P` at `Q` using a spherical mirror, with `XY` as the optic axis. The mirror must beA. convergingB. divergingC. positoined to the left of `P`D. positioned to the right of `Q`
Answer» Correct Answer - A::C

Discussion

791.	A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C be its centre of curvature. A point object is placed at C, whose real image is also formed at C. If the mirror is now filled with water, the image will be (a) real, and will remain at C (b) real, and will be located above C (c) virtual, and will be located below O (d) real, and will be located between C and O
Answer» Correct Answer is: (d) real, and will be located between C and O When the mirror is filled with water, the apparent position of the object, as seen from the pole of the mirror, will be above C. Hence, the image will be formed below C, i.e., between C and O.

791.

A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C be its centre of curvature. A point object is placed at C, whose real image is also formed at C. If the mirror is now filled with water, the image will be (a) real, and will remain at C (b) real, and will be located above C (c) virtual, and will be located below O (d) real, and will be located between C and O

Answer»

Correct Answer is: (d) real, and will be located between C and O

When the mirror is filled with water, the apparent position of the object, as seen from the pole of the mirror, will be above C. Hence, the image will be formed below C, i.e., between C and O.

Discussion

792.	Two points P and Q lie on either side of an axis XY as shown. It is desired to produce an image of P at Q using a spherical mirror, with XY as the optic axis. The mirror must be (a) converging (b) diverging (c) positioned to the left of P (d) positioned to the right of Q
Answer» Correct Answer is: (a) converging , (c) positioned to the left of P

Discussion

793.	An object and a screen are fixed at a distance d apart. When a lens of focal length f is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are M1 and M2.(a) d > 2f(b) d > 4f(c) M1M2 = 1d) \|M1\| - \|M2\| = 1
Answer» Correct Answer is: (b) d > 4f , (c) M₁M₂ = 1

Discussion

794.	Two periodic waves of intensities `I_(1)` and `I_(2)` pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:A. `I_(1) + I_(2)`B. `(sqrt(I_(1)) + sqrt(I_(2)))^(2)`C. (c ) `(sqrt(I_(1)) - sqrt(I_(2)))^(2)`D. `2(I_(1) + I_(2))`
Answer» Correct Answer - D In interference, resultant intensity is `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 0^(@)` `= I_(1) + I_(2) + 2sqrt(I_(1)I_(2))` `I_(min) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 180^(@)` `= I_(1) + I_(2) - 2sqrt(I_(1)I_(2))` `I_(max) + I_(min) = 2 (I_(1) + I_(2))`

Discussion

795.	A beam of electron is used `YDSE` experiment . The slit width is d when the velocity of electron is increased ,thenA. no interference is observedB. fringe width increasesC. fringe width decreasesD. fringe width remains same
Answer» Correct Answer - C When velocity of electron increases, the momentum of electron `(p = mv)` increases. The de-Broglic wavelength `lambda(h//mv)` decreases. The fringe width is `beta = lambda D//d`, so `beta prop lambda`. Hence if `lambda` decreases, `beta` decreases.

Discussion

796.	State the conditions under which a rainbow can be seen.
Answer» A rainbow can be observed when there is a light shower with relatively large raindrop occurring during morning or evening time with enough sunlight around.

Discussion

797.	The intensity ratio in the interference pattern is `1 : 9`. What is the amplitude ratio and the ratio of widths of two slits ?
Answer» Correct Answer - `2 : 1 ; 4 : 1` `(I_(min))/(I_(max)) = ((a - b)^(2))/((a + b)^(2)) = (1)/(9)` `:. (a - b)/(a + b) = (1)/(3) or 3 a - 3 b = a + b` `2 a = 4b or a = 2 b` Amplitude ratio : `(a)/(b) = (2)/(1)` `(w_(1))/(w_(2)) = (I_(1))/(I_(2)) = (a^(2))/(b^(2)) = ((2)/(1))^(2) = 4 : 1`

Discussion

798.	An astronomical telescope in normal adjustment receives light from a distant source S. The tube length is now decreased slightly. (a) A virtual image of S will be formed at a finite distance. (b) No image will be formed.(c) A small, real image of S will be formed behind the eyepiece, close to it. (d) A large, real image of S will be formed behind the eyepiece, far away from it.
Answer» Correct Answer is: (a) When tube length is decreased, the (real) intermediate image formed by the objective will lie between the eyepiece and its focus. This will cause a virtual image to be formed.

798.

An astronomical telescope in normal adjustment receives light from a distant source S. The tube length is now decreased slightly. (a) A virtual image of S will be formed at a finite distance. (b) No image will be formed.(c) A small, real image of S will be formed behind the eyepiece, close to it. (d) A large, real image of S will be formed behind the eyepiece, far away from it.

Answer»

Correct Answer is: (a)

When tube length is decreased, the (real) intermediate image formed by the objective will lie between the eyepiece and its focus. This will cause a virtual image to be formed.

Discussion

799.	An astronomical telescope in normal adjustment receives light from a distant source S, if the tube length is increased slightly from its position of normal adjustment, (a) a virtual image of S will be formed at a finite distance (b) no image will be formed (c) a small, real image of S will be formed behind the eyepiece, close to it (d) a large, real image of S will be formed behind the eyepiece, far away from it
Answer» Correct Answer is:(d) When tube length is increased, the (real) intermediate image formed by the objective will lie beyond the focus of the eyepiece. This will produce a real, magnified image behind the eyepiece.

799.

An astronomical telescope in normal adjustment receives light from a distant source S, if the tube length is increased slightly from its position of normal adjustment, (a) a virtual image of S will be formed at a finite distance (b) no image will be formed (c) a small, real image of S will be formed behind the eyepiece, close to it (d) a large, real image of S will be formed behind the eyepiece, far away from it

Answer»

Correct Answer is:(d)

When tube length is increased, the (real) intermediate image formed by the objective will lie beyond the focus of the eyepiece. This will produce a real, magnified image behind the eyepiece.

Discussion

800.

In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is(a) L/l(b) L/l + 1 (c) L/l - 1(d) L + l / L - l

Answer»

Correct Answer is: (a) L/l

Let f_o and f_e be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f_o + f_e. Treating the line on the objective as the object, and the eyepiece as the lens, u = -(f_o + f_e) and f = f_e.

1/v - 1/-(f_o + f_e) = 1/f_e

or 1/v = 1/f_e - 1/f_o + f_e = f_o/(f_o + f_e) f_e

or v = (f_o + f_e)f_e / f_o

Magnification =|v/u| = f_e/f_o = image size/object size = l/L:

∴ f_o/f_e = L/l = magnification of telescope in normal adjustment.

Discussion

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