Q: Q.5.206. A sounding of dilute plasma by radiowaves of various frequencies reveals that radiowaves with wavelengths exceeding λo = 0.75 m experience total internal reflection. Find the free electron concentration in that plasma.

0=1−ε0mω2n0e2 if ω=λ02πc Hence 4π2ε0mc2n0e2λ02=1 or n0=e2λ024π2E0mc2=1.984×109 per c.c.

Q: Q.18. A man standing on the edge of a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is equal to h. At what distance from the surface of water is the image of the stone formed if the line of vision makes an angle θ with the normal to the surface?

From the Fig.sindα=OMMP=hsec(α+dα)MNcosα As dα is very small, sodα≈hsecαMNcosα=hMNcos2α (1)Similarlydθ=h′MNcos2θ (2)From Eqns (1) and (2)dθdα=hcos2θh′cos2α or ,h′=cos2αhcos2θdθdα (3)From the law of refractionnsinα=sinθ (A)sinα=nsinθ so, cosα=n2n2−sin2θ (B)Differentiating Eqn.(A)ncosαdα=cosθdθ or, dθdα=ncosαcosθ (4)Using (4) in (3), we geth′=ncos3αhcos3θ Hence, h′=n(n2n2−sin2θ)3/2hcos3θ=(n2−sin2θ)3/2n2hcos3θ [Using Eqn.(B)] (5)

Q: Q.21. The least deflection angle of a certain glass prism is equal to its refracting angle. Find the latter.

In this case we havesin2α+θ=nsin2θ α=θ So, sinθ=nsin(θ/2) or, 2sin(θ/2)cos(θ/2)=nsin(θ/2) Hence, cos(θ/2)=2n or, θ=2cos−1(n/2)=83∘ where, n=1.5

Question 1

Q.10. A narrow beam of alpha particles with kinetic energy T = = 0.50 MeV and intensity I = 5.0.10^5  particles per second falls normally on a golden foil. Find the thickness of the foil if at a distance r = 15 cm from a scattering section of that foil the flux density of scattered particles at the angle θ = 60° to the incident beam is equal to J = 40 particles/(cm^2•s).

Accepted Answer

A scattered flux density of J (particles per unit area per second) equals  J/r21​=r2J particles scattered per unit time per steradian in the given direction. Let n = concentration of the gold nuclei in the foil. Thenn=AAu​NA​ρ​ and the number of Au nuclei per unit area of the foil is nd where d = thickness of the foil Then from Eqn. (6.1 b) (note that n → nd here)r2J=dN=ndI((4πε0​)2TZe2​)2cosec42θ​ Here I  is the number of α - particles falling on the foil per secondHence   d=nI(4πε0​Ze2​)24T2r2J​sin4θ/2 using Z = 79 , AAu =197,ρ=19−3×103 kg/m3 NA​=6.023×1026 / kilo mole and other data from the problem we getd = 1.47 pm

Question 2

Q.4. Determine the luminosity of a surface whose luminance depends on direction as L = Lo  cosθ, where θ is the angle between the radiation direction and the normal to the surface.

Accepted Answer

Luminance L is the light energy emitted per unit area of the emitting surface in a given direction per unit solid angle divided by cosθ. Luminosity M is simply energy emitted per unit area. ThusM=∫L⋅cosθ⋅dΩ where the integration must be in the forward hemisphere of the emitting surface (assuming light is being emitted in only one direction say outward direction of the surface.) ButThusL=L0​cosθ M=∫L0​cos2θ⋅dΩ=2π∫0π/2​L0​cos2θsinθdθ=32​πL0​

Question 3

Q.5. A certain luminous surface obeys Lambert's law. Its luminance is equal to L. Find: (a) the luminous flux emitted by an element ΔS of this surface into a cone whose axis is normal to the given element and whose aperture angle is equal to θ; (b) the luminosity of such a source.

Accepted Answer

(a) For a Lambert source L = ConstThe flux emitted into the cone isΦ=LΔScosαdΩ =LΔS∫0θ​2πcosαsinαdα =LΔSπ(1−cos2θ)=πLΔSsin2θ (b) The luminosity is obtained from the previous formula for θ=90∘ M=ΔSΦ(θ=90∘)​=πL

Question 4

Q.5.206. A sounding of dilute plasma by radiowaves of various frequencies reveals that radiowaves with wavelengths exceeding λo = 0.75 m experience total internal reflection. Find the free electron concentration in that plasma.

Accepted Answer

0=1−ε0​mω2n0​e2​ if ω=λ0​2πc​ Hence 4π2ε0​mc2n0​e2λ02​​=1 or n0​=e2λ02​4π2E0​mc2​=1.984×109 per c.c.

Question 5

Q.15. At what value of the angle of incident θ1 is a shaft of light reflected from the surface of water perpendicular to the refracted shaft?

Accepted Answer

Let PQ be the surface of water and n be the R.I. of water. Let AO is the shaft of light with incident angle θ1 and OB and OC are the reflected and refracted light rays at angles θ1​ and θ2​ respectively (Fig.). From the figure θ2​=2π​−θ1​ From the law of refraction at the interface PQ n=sinθ2​sinθ1​​=sin(2π​−θ1​)sinθ1​​ or, n=cosθ1​sinθ1​​=tanθ1​ Hence, θ1​=tan−1n

Question 6

Q.16. Two optical media have a plane boundary between them. Supposeθicr is the critical angle of incidence of a beam and θ1  is the angle of incidence at which the refracted beam is perpendicular to the reflected one (the beam is assumed to come from an optically denser medium). Find the relative refractive index of these media if sinθ1cr​/sinθ1​=η=1.28

Accepted Answer

Let two optical mediums of R.I.n1 and n2 respectively be such that n1 >n2 . In the case when angle of incidence is θ1cr​ (Fig.), from the law of refraction n1​sinθ1cr​=n2​               (1)In the case , when the angle of incidence is θ1, from the law of refraction at the interface of mediums 1 and 2 .n1​sinθ1​=n2​sinθ2​ But in accordance with the problem θ2​=(π/2−θ1​) so,        n1​sinθ1​=n2​cosθ2​          (2)Dividing Eqn (1) by (2)sinθ1​sinθ1​a​=cosθ1​1​ or,       η=cosθ1​1​ so, cosθ1​=η1​ and sinθ1​=ηη2−1​​ (3)But    n2​n1​​=sinθ1​cosθ1​​ So,    n2​n1​​=η1​η2−1​η​     (Using 3)Thus, n2​n1​​=η2−1​1​

Question 7

Q.17. A light beam falls upon a plane-parallel glass plate d=6.0 cm in thickness. The angle of incidence is θ = 60°. Find the value of deflection of the beam which passed through that plate.

Accepted Answer

From the Fig. the sought lateral shift x=OMsin(θ−β) =dsecβsin(θ−β) =dsecβ(sinθcosβ−cosθsinβ) =d(sinθ−cosθtanβ) (1)But from the law of refractionsinθ=nsinβ or, sinβ=nsinθ​ So, cosβ=nn2−sin2θ​​ and, tanβ=n2−sin2θ​sinθ​ x=d(sinθ−cosθtanβ)=d(sinθ−cosθn2−sin2θ​sinθ​) Thus, =dsinθ[1−n2−sin2θ1−sin2θ​​]

Question 8

Q.18. A man standing on the edge of a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is equal to h. At what distance from the surface of water is the image of the stone formed if the line of vision makes an angle θ with the normal to the surface?

Accepted Answer

From the Fig.sindα=OMMP​=hsec(α+dα)MNcosα​ As dα is very small, sodα≈hsecαMNcosα​=hMNcos2α​ (1)Similarlydθ=h′MNcos2θ​ (2)From Eqns (1) and (2)dθdα​=hcos2θh′cos2α​ or ,h′=cos2αhcos2θ​dθdα​          (3)From the law of refractionnsinα=sinθ           (A)sinα=nsinθ​ so, cosα=n2n2−sin2θ​​            (B)Differentiating Eqn.(A)ncosαdα=cosθdθ or, dθdα​=ncosαcosθ​                   (4)Using (4) in (3), we geth′=ncos3αhcos3θ​ Hence, h′=n(n2n2−sin2θ​)3/2hcos3θ​=(n2−sin2θ)3/2n2hcos3θ​ [Using Eqn.(B)]          (5)

Question 9

Q.19. Demonstrate that in a prism with small refracting angle θ the shaft of light deviates through the angle α≃(n−1)θ regardless of the angle of incidence, provided that the latter is also small.

Accepted Answer

The figure shows the passage of a monochromatic ray through the given prism, placed in air medium.From the figure, we haveθ=β1​+β2​ (A)and, α=(α1​+α2​)−(β1​+β2​) α=(α1​+α2​)−θ (1)From the Snell’s lawsinα1​=πsinβ1​ or, α1​=nβ1​ (for small angles) (2)and, sinα2​=nsinβ2​ or, α2​=nβ2​ (for small angles) (3)From Eqns (1), (2) and (3), we getα=n(β1​+β2​)−θ So, α=n(θ)−θ=(n−1)θ  [Using Eqn.A]

Question 10

Q.21. The least deflection angle of a certain glass prism is equal to its refracting angle. Find the latter.

Accepted Answer

In this case we havesin2α+θ​=nsin2θ​ α=θ So, sinθ=nsin(θ/2) or, 2sin(θ/2)cos(θ/2)=nsin(θ/2) Hence, cos(θ/2)=2n​ or, θ=2cos−1(n/2)=83∘ where, n=1.5

Question 11

Q.22. Find the minimum and maximum deflection angles for a light ray passing through a glass prism with refracting angle θ = 60°.

Accepted Answer

In the case of minimum deviation,sin2α+θ​=nsin2θ​  So, α=2sin−1{nsin2θ​}−θ=37∘ for, n=1.5 Passage of ray for grazing incidence and grazing emergence is the condition for maximum deviation (Fig.). From Fig.α=π−θ=π−2θcr​ (where θcr​ is the critical angle)So, α=π−2sin(1/n)=58∘ for n=1.5= R.I. of glass.

Question 12

Q.23. A trihedral prism with refracting angle 60° provides the least deflection angle 37° in air. Find the least deflection angle of that prism in water.

Accepted Answer

The least deflection angle is given by the formula, δ=2α−θ ,where α is the angle of incidence at first surface and θ is the prism angle.Also from Snell’s law,n1​sinαx=n2​sin(θ/2) as the angle of refraction at first surface is equal to half the angle of prism for least deflectionso, sinα=n1​n2​​sin(θ/2)=1.331.5​sin30∘=.5639 or, α=sin−1(.5639)=34.3259∘ Substituting in the above (1), we get,δ=8⋅65∘

Question 13

Q.24. A light ray composed of two monochromatic components passes through a trihedral prism with refracting angle θ = 60°. Find the angle Δα between the components of the ray after its passage through the prism if their respective indices of refraction are equal to 1.515 and 1.520. The prism is oriented to provide the least deflection angle.

Accepted Answer

From the Cauchy’s formula, and also experimentally the R.I. of a medium depends upon the wavelength of the monochromatic ray i.e.n=f(λ) In the case of least deviation of a monochromatic ray the passage a prism, we have:  nsin2θ​=sin2α+θ​  (1)The above equation tales us that we have n = n (α) , so we may writeΔn=dαdn​Δα   (2)From Eqn. (1)dnsin2θ​=21​cos2α+θ​dα or dαdn​=2sin2θ​cos2α+θ​​   (3)From Eqn. (2) and (3)Δn=2sin2θ​cos2α+θ​​Δα Δn=2sin2θ​1−sin2(2a+θ​)​​Δα=2sin2θ​1−n2sin22θ​​​Δα Thus  Δα=1−n2sin22θ​​2sin2θ​​Δn=0.44

Question 14

Q.25. Using Fermat's principle derive the laws of deflection and refraction of light on the plane interface between two media.

Accepted Answer

Ans. Fermat’s principle : “ The actual path of propagation of light (trajectory of a light ray ) is the path which can be followed by light with in the lest time, in comparison with all other hypothetical paths between the same two points. ” “Above statement is the original wordings of Fermat ( A famous French scientist of 17th century)” Deduction of the law of refraction from Fermat’s principle : Let the plane S be the interface between medium 1 and medium 2 with the refractive indices n1​=c/v1​ and n2​=c/v2​  Fig.(a). Assume, as usual, that n12 . Two points are given - one above the plane S (point A ), the other under plane S (point B ). The various distances are :AA1​=h1​,BB1​=h2​,A1​B1​=l. We must find the path from A to B which can be covered by light faster than it can cover any other hypothetical path. Clearly, this path must consist of two straight lines, viz, AO in medium 1 and OB in medium 2; the point O in the plane S has to be found.First of all, it follows from Fermat’s principle that the point O must lie on the intersection of S and a plane P, which is perpendicular to S and passes through A and B.Indeed, let us assume that this point does not lie in the plane P; let this be point 0 1 in Fig. (b). Drop the perpendicular O2 from O1 onto P . Since AO2 1 and BO21 it is clear that the time required to traverse AO2B is less than that needed to cover the path AO1B.Thus, using Fermat’s principle, we see that the first law of refraction is observed : the incident and the refracted rays lie in the same plane as the perpendicular to the interface at the pointwhere the ray is refracted. This plane is the plane P in Fig. (b); it is called the plane of incidence.Now let us consider light rays in the plane of incidence Fig. (c). DesignateA1​O as x and OB1​=1−x The time it takes a ray to travel from A to O and then from O to B isT=v1​AO​+v2​OB​=v1​h12​+x2​​+v2​h22​+(l−x)2​​     (1)The time depends on the value of x. According to Fermat's principle, the value of x must m inim ize the time T. A t this value of x the derivative dT/dx equals zero :dxdT​=v1​h12​+x2​x​−v2​h22​+(l−x)2​l−x​=0     (2)Now,h12​+x2​x​=sinα, and h22​+(l−x)2​l−x​=sinβ Consequently,v1​sinα​−v2​sinβ​=0, or sinβsinα​=v2​v1​​ So,sinβsinα​=c/n2​c/n1​​=n1​n2​​ Note : Fermat himself could not use Eqn. 2. as mathematical analysis was developed later by Newton and Leibniz. To deduce the law of the refraction of light, Fermat used his own maximum and minimum method of calculus, which, in fact, corresponded to the subsequently developed method of finding the minimum (maximum) of a function by differentiating it and equating the derivative to zero.

Question 15

Q.26. By means of plotting find: (a) the path of a light ray after reflection from a concave and convex spherical mirrors (see Fig. 5.4, where F is the focal point, 00' is the optical axis); (b) the positions of the mirror and its focal point in the cases illustrated in Fig. 5.5, where P and P' are the conjugate points.

Accepted Answer

Ans.  (a) Look for a point O' on the axis such that O' F and O' P make equal angles with O' O.This determines the position of the mirror. Draw a ray from P parallel to the axis. This must on reflection pass through F. The intersection of the reflected ray with principal axis determines the focus.

(b) Suppose P is the object and F is the image. Then the mirror is convex because the image is virtual, erect & dim inished. Look for a point X (between P & F ) on the axis such that PX and F'X make equal angle with the axis.

Question 16

Q.28. A point source with luminous intensity I0  = 100 cd is positioned at a distance s = 20.0 cm from the crest of a concave mirror with focal length f = 25.0 cm. Find the luminous intensity of the reflected ray if the reflection coefficient of the mirror is p = 0.80.

Accepted Answer

For a concave mirror as usual s′1​+s1​=f1​ So, s′=s−fsf​ (In coordinate convention s = - s is negative & f=−∣f∣ is also negative.)If A is the area of the mirror (assumed sm all) and the object is on the principal axis, then the light incident on the mirror per second is I0​s2A​ This follows from the definition of luminous intensity as light emitted per second per unit solid angle in a given direction and the fact that s2A​ is the solid angle subtended by the mirror at the source. of this a fraction p is reflected so if I is the luminous intensity of the image, then Is2A​=ρI0​s2A​ Hence,I=ρI0​(∣f∣−s∣f∣​)2 (Because our convention makes f - ve for a concave mirror, we have to write ∣f∣ )Substitution gives I=2.0×103 cd

Question 17

Q.29. Proceeding from Fermat's principle derive the refraction formula for paraxial rays on a spherical boundary surface of radius R between media with refractive indices n and n'.

Accepted Answer

Ans. For O1 to be the image, the optical paths of all rays OAO1 must be equal upto terms o f leading order in h Thusn1​OA+n2​AO1​= constantSo, OP=∣s∣,O1​P=∣s′∣ and soOA=h2+(∣s∣+δ)2​≈∣s∣+δ+2∣s∣h2​ O1​A=h2+(∣s′∣−δ)2​≈∣s′∣−δ+2∣s′∣h2​ (neglecting products h2δ ). Thenn1​∣s∣+n2​∣s′∣+n1​δ−n2​δ+2h2​(∣s∣n1​​+∣s′∣n2​​)=Const. Now, (r−δ)2+h2=r2 or, h2=2r˙ or, δ=2rh2​ Here, r=CP Hence, n1​∣s∣+n2​∣s′∣+2h2​{rn1​−n2​​+∣s∣n1​​+∣s′∣n2​​}= ConstantSince this must hold for all h, we have∣s′∣n2​​+∣s∣n1​​=rn2​−n1​​ From our sign convention,s′>0,s<0 so we get,s′n2​​−sn1​​=rn2​−n1​​.

Question 18

Q.30. A parallel beam of light falls from vacuum on a surface enclosing a medium with refractive index n (Fig. 5.6). Find the shape of that surface, x (r), if the beam is brought intofocus at the point F at a distance f from the crest O. What is the maximum radius of a beam that can still be focussed?

Accepted Answer

Ans. All rays focusing at a point must have traversed the same optical path. Thus x+nr2+(f−x)2​=nf or, (nf−x)2=n2r2+n2(f−x)2 or, n2r2=(nf−x)2−[n(f−x)]2=(nf−x+nf−nx)(nf−x−nf+nx) =x(n−1)(2nf−(n+1)x) =2n(n−1)fx−(n+1)(n−1)x2 Thus, (n+1)(n−1)x2−2n(n−1)fx+n2r2=0 So, x=(n+1)(n−1)n(n−1)f±n2(n−1)2f2−n2r2(n+1)(n−1)​​ =n+1nf​[1±1−n−1n+1​f2r2​​] Ray must move forward so x<f, for +sign x>f for small r so -sign.(Also x→0 as r→0 )(x>f means ray turning back in the direction of incidence. (see Fig.)) Hence, x=n+1nf​[1−1−n−1n+1​f2r2​​] For the maximum value of r 1−n−1n+1​f2r2​​=0         (A)because the expression under the radical sign must be non-negative, which gives the maximum value of r.Hence from Eqn. (A),rmax​=f(n−1)/(n+1)​

Question 19

Q.31. A point source is located at a distance of 20 cm from the front surface of a symmetrical glass biconvex lens. The lens is 5.0 cm thick and the curvature radius of its surfaces is 5.0 cm. How far beyond the rear surface of this lens is the image of the source formed?

Accepted Answer

As the given lens has significant thickness, the thin lens, formula cannot be used.For refraction at the front surface from the formula s′n′​−sn​=Rn′−n​ s′1⋅5​−−201​=51⋅5−1​ On simplifying we get , s' = 30 cm.Thus the image l' produced by the front surface behaves as a virtual source for the rear surface at distance 25 cm from it, because the thickness of the lens is 5 cm. Again from the refraction formula at curve surfaces′n′​−sn​=Rn′−n​ s′1​−251⋅5​=−51−1.5​ On simplifying, s′=+6.25 cm Thus we get a real image l at a distance 6- 25 cm beyond the rear surface (Fig.).

Question 20

Q.32. An object is placed in front of convex surface of a glass piano-convex lens of thickness d = 9.0 cm. The image of that object is formed on the plane surface of the lens serving as'a screen. Find: (a) the transverse magnification if the curvature radius of the lens's convex surface is R = 2.5 cm; (b) the image illuminance if the luminance of the object is L = 7700cd/m^2  and the entrance aperture diameter of the lens is D = 5.0 mm; losses of light are negligible.

Accepted Answer

(a) The formation of the image of a source 5, placed at a distance u from the pole of the convex surface of plano-convex lens of thickness d is shown in the figure.On applying the formula for refraction through spherical surface, we gets′n​−s1​=(n−1)/R (here, n2​=n and n1​=1 )or, dn​−s1​=(n−1)/R s1​=dn​−R(n−1)​ or, s1​=dn​−R(n−1)​ or, ss′​=s′{dn​−R(n−1)​} But in this case optical path of the light, corresponding to the distance v in the medium is v/n, so the magnification produced will be,β=nss′​=ns′​{dn​−R(n−1)​}=nd​{dn​−R(n−1)​}=1−nRd(n−1)​ Substituting the values, we get magnification β = - 0-20.(b) If the transverse area of the object is A (assumed small), the area of the image is  β^2A.We shall assume that 4πD2​>A. Then light falling on the Jens is LAs2πD2/4​ from the definition of luminance cosθ∼1 if D2≪s2 and dΩ=s2πD2/4​ Then the illuminance of the image isLAs2πD2/4​/β2A=Ln2πD2/4d2 Substitution gives 42 lx.

Chapter 5 : PART FIVE. OPTICS