Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

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: Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

The document Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev is a part of the Course Physics For JEE.

Q.41. Determine the focal length of a concave spherical mirror which is manufactured in the form of a thin symmetric biconvex glass lens one of whose surfaces is silvered. The curvature radius of the lens surface is R = 40 cm. 

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Ans.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.42. Figure 5.9 illustrates an aligned system consisting of three thin lenses. The system is located in air. Determine:
 (a) the position of the pointof convergence of a parallel ray incoming from the left after passing through the system;
 (b) the distance between the first lens and a point lying on the axis to the leftof the system, at which that point and its image are located symmetrically with respect to the lens system. 

Ans.  ( a ) Path of a ray, as it passes th rough the lens system is as shown below.
F o c a l le n g th of all the three lenses,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev  , n e g le c t in g their signs.

Applying lens formula for the first lens, considering a ray coming from infinity,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and so the position of the image is 5 cm to the right of the second lens, when only the firstone is present, but the ray again gets refracted while passing through the second, so

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

or, s' = 10 cm, which is now 5 cm left to the third lens sofor this lens,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

or,Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(b) This means that if the object is x cm to be left of the first lens on the axis OO' then the image is x on to the right of the 3rd (la s t ) lens. Call the lenses 1,2,3 from the left and leto be the object, O1 its image by the first lens, O2 the image of O1 by the 2nd lens and O3, the image of O2 by the third lens.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

O1 and O2 must be symmetrically located with respect to the lens Land since this lens is concave, O1 must be at a distance Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev to be the right of L2 and O2 must be Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev to be the left of L2, One can check that this satisfies lens equation for the third lens Ls

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

HenceIrodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.43. A Galilean telescope of 10-fold magnification has the length of 45 cm when adjusted to infinity. Determine:
 (a) the focal lengths of the telescope's objective and ocular;
 (b) by what distance the ocular should be displaced to adjust the telescope to the distance of 50 m. 

Ans. (a) Angular magnification for Galilean telescope in normal adjustment is given as.

 Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

or,  Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev  (1)

The length of the telescope in this case.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

So, using (1), we get,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.44. Find the magnification of a Keplerian telescope adjusted to infinity if the mounting of the objective has a diameter D and the image of that mounting formed by the telescope's ocular has a diameter d.

Ans. In the Keplerian telescope, in normal adjustment, the distance between the objective and eyepiece is f0 + fe . The image of the mounting produced by the eyepiece is formed at a distance v to the right where

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

But  Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

soIrodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

The linear magnification produced by the eyepiece of the mounting is, in magnitude,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

This equals Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev according to the problem so

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.45. On passing through a telescope a flux of light increases its intensity η = 4.0.104  times. Find the angular dimension of a distantobject if its image formed by that telescope has an angular dimension ψ' = 2.0°. 

Ans. It is clear from the figure that a parallel beam of light, originally of intensity l0 has, on emerging from the telescope, an intensity.
Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

because it is concentrated over a section whose diameter is fe/fo of the diameter of the cross section of the incident beam.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.46. A Keplerian telescope with magnification Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev = 15 was submerged into water which filled up the inside of the telescope. To make the system work as a telescope again within the former dimensions, the objective was replaced.What has the magnification of the telescope become equal to? The refractive index of the glass of which the ocular is made is equal to n == 1.50. 

Ans. W hen a glass lens is immersed in w ater its focal length increases approximately four times. We check this as follows as :

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now back to the problem. Originally in air

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

In water, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and the focal length of the replaced objective is given by the condition

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.47. At what magnification Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev of a telescope with a diameter of the objective D = 6.0 cm is the illuminance of the image of an objecton the retina not less than without the telescope? The pupil diameter is assumed to be equal to d0 = 3.0 mm. The losses of light in the telescope are negligible. 

Ans. If L is the luminance of the object, A is its area, s = distance of the object then light falling on the objective is

 Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

The area of the image formed by the telescope (assuming that the image coincides with the object) is Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev  A and the area of the final image on the retina is

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Where f = focal length of the eye lens. Thus the illuminance of the image on the retin (when the object is observed through the telescope) is

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

When the object is viewed directly, the illuminance is, similarly, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

We want Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

So,Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev on substitution of the values.

 

Q.48. The optical powers of the objective and the ocular of a microscope are equal to 100 and 20 D respectively. The microscope magnification is equal to 50. What will the magnification of the microscope be when the distance between the objective and the ocular is increased by 2.0 cm? 

Ans. Obviously, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now, we know that, magnification of a microscope,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev  for distinct vision

or,  Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Since distance between objective and ocular has increased by 2 cm, hence it will cause the increase of tube length by 2cm.

so, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and hence, :Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.49. A microscope has a numerical aperture sin α = 0.12, where a is the aperture angle subtended by the entrance pupil of the microscope. Assuming the diameter of an eye's pupil to be equal to d= = 4.0 mm, determine the microscope magnification at which
 (a) the diameter of the beam of light coming from the microscope is equal to the diameter of the eye's pupil;
 (b) the illuminance of the image on the retina is independentof magnification (consider the case when the beam of light passing through the system "microscope-eye" is bounded by the mounting of the objective).

Ans. It is implied in the problem that final image of the object is at infinity (otherwise light coming outof the eyepiece will not have a definite diameter).

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

But when the final image is at infinity, the magnification Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev in a microscope is given by

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(b) If Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev is the magnification produced by the microscope, then the area of the image produced on the retina (when we observe an object through a microscope) is:

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Where u = distance of the image produced by the microscope from the eye lens, f = focal length of the eye lens and A = area of the object If  φ = luminous flux reaching the objective from the object and d = dso that the entire flux is admitted into the eye), then the illuminance of the final image on the retinaIrodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

But if Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRevthen only a fraction (d0 | d)2  of light is admitted into the eye and the illuminance becomes

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

independentof Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev The condition for this is then

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.50. Find the positions of the principal planes, the focal and nodal points of a thin biconvex symmetric glass lens with curvature radius of its surfaces equal to R = 7.50 cm. There is air on one side of the lens and water on the other.

Ans. The primary and secondary focal length of a thick lens are given as,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

where φ is the lens power n , n' and n" are the refractive indices of first medium, lens material and the second medium beyond the lens.Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev are the powers of first and second spherical surface of the lens.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev       (1)

Now, power of a thin lens,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev          (2)

From equations (1) and (2), we get

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and    Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Since the distance between the primary principal point and primary nodal point is given as,

So, in this case,     Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.51. By means of plotting find the positions of focal points and principal planes of aligned optical systems illustrated in Fig. 5.10: 

(a) a telephoto lens, that is a combination of a converging and a diverging thin lenses (f1  = 1.5 a, f2  = —1.5 a); 

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(b) a system of two thin converging lenses  Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev
 (c) a thick convex-concave lens Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Ans.

  

Q.52. An optical system is located in air. Let 00' be its optical axis, F and F' are the front and rear focal points, H and H' are the front and rear principal planes, P and P' are the conjugate points. By means of plotting find:
 (a) the positions F' and H' (Fig. 5.11a);
 (b) the position of the point S' conjugate to the point S (Fig. 5.11b); 

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(c) the positions F, F', and H' (Fig. 5.11c, where the path of the ray of light is shown before and after passing through the system). 

Ans. (a) Draw P' X parallel to the axis OO' and let PF interest it at X. That determines the principal point H. As the' medium on both sides of the system is the same, the principal point coincides with the nodal point Draw a ray parallel to PH through P' . That determines H' Draw a ray P X' parallel to the axis and join P'X".
That gives F' .

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(b) We let H stand for the principal point (on the axis). Determ ine H" by drawing a ray P'H" passing through P' and parallel to PH. One ray (conjugate to SH) can be obtained from this. To get the other ray one needs to know F or F" . This is easy because P and F are known. Finally we get S'.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(c) From the incident ray we determine Q. A line parallel to OO' through Q determines Q and hence H'.
H and H ' are then also the nodal points. A ray p arallel to the incident ray through H will emeige parallel to itse lf through H '. That determines F . Sim ilatiy a ray parallel to the emergent ray through H determines F.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.53. Suppose F and F' are the front and rear focal points of an optical system, and H and H' are its front and rear principal points. By means of plotting find the position of the image S' of the point S for the following relative positions of the points S, F, F', H, and H':

(a) FSHH'F'; (b) HSF'FH'; (c) H'SF'FH; (d) F'H'SHF. 

Ans.  Here we do not assume that the media on the two sides of the system are the same.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.54. A telephoto lens consists of two thin lenses, the front converging lens and the rear diverging lens with optical powers Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRevIrodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRevFind:
 (a) the focal length and the positions of principal axes of that system if the lenses are separated by a distance d = 4.0 cm;
 (b) the distance d between the lenses at which the ratio of a focal length f of the system to a distance l between the converging lens and the rear principal focal point is the highest. What is this ratio equal to? 

Ans. (a) Optical power of the system of combination of two lenses,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

on putting the values,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

or,  Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now, the position of primary principal plane with respect to the vertex of conveiging lens,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Similarly, the distance of secondary principal plane with respect to the vertex of diverging lens.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

(b) The distance between the rear principal focal point F' and the vertex of conveiging lens,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now , if f/l is maximum for certain value of d then f/l will be minimum for the same value of d. And for minimum f/l,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.55. Calculate the positions of the principal planes and focal points of a thick convex-concave glass lens if the curvature radius of the convex surface is equal to R1  = 10.0 cm and of the concave surface to R2 = 5.0 cm and the lens thickness is d = 3.0 cm. 

Ans. The optical power of first convex surface is,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and the optical power of second concave surface is,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

So, the optical power of the system,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now , the distance of the primary principal plane from the vertex of convex surface is given as,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and the distance of secondary principal plane from the vertex of second concave surface,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.56. An aligned optical system consists of two thin lenses with focal lengths f1  and f2 the distance between the lenses being equal to d. The given system has to be replaced by one thin lens which, at any position of an object, would provide the same transverse magnification as the system. What must the focal length of this lens be equal to and in what position must it be placed with respect to the two-lens system? 

Ans. The optical power of the system of two thin lenses placed in air is given as, 

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev       (1)

This equivalent fo cal length of the'system of two lenses is measured from the primary principal plane.
A s clear from the figure, the distance of the primary principal plane from the optical centre of the first is

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now, if we place the equivalent lens at the primary principal plane of the lens system, it will provid e the same transverse magnification as the system. So, the distance of equivalent lens from the Vertex of the first lens is,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.57. A system consists of a thin symmetrical converging glass lens with the curvature radius of its surfaces R = 38 cm and a plane mirror oriented at right angles to the optical axis of the lens. The distance between the lens and the mirror is l = 12 cm. What is the optical power of this system when the space between the lens and the mirror is filled up with water? 

Ans. The plane mirror forms the image of the lens, and water, filled in the space between the two, behind the mirror, as shown in the figure.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

= optical power of individual lens and n0 = R.I. of water.

Now, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev optical power of first convex surface + optical power of second concave surface.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRevn is the refractive index of glass.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev(1)

and so, the optical power of whole system,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRevsubstituting the values.

 

Q.58. At what thickness will a thick convex-concave glass lens in air
 (a) serve as a telescope provided the curvature radius of its convex surface is ΔR = 1.5 cm greater than thatof its concave surface?
 (b) have the optical power equal to —1.0 D if the curvature radii of its convex and concave surfaces are equal to 10.0 and 7.5 cm respectively? 

Ans. (a) A telescope in normal adjustment is a zero power conbination of lenses. Thus we require

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev
 

Q.59. Find the positions of the principal planes, the focal length and the sign of the optical power of a thick convex-concave glass lens
 (a) whose thickness is equal to d and curvature radii of the surfaces are the same and equal to R;
 (b) whose refractive surfaces are concentric and have the curvature radii R1  and R2 (R2> R1). 

Ans. (a) The power of the lens is (as in the previous problem)

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

The principal planes are located on the side of the convex surface at a distance d from each other, with the front principal plane being removed from the con vex surface of the lens by a distance R /(n - 1) .

(b)

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.60. A telescope system consists of two glass balls with radii R= 5.0 cm and R2 = 1.0 cm. What are the distance between the centres of the balls and the magnification of the system if the bigger ball serves as an objective?

Ans. Let the optical powers of the first and second surfaces of the ball of radius R1 beIrodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev then

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

This ball may be treated as a thick spherical lens of thickness 2R1. So the optical power of this sphere is,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev        (1)

Similarly, the optical power of second ball,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

If the distance between the centres of these balls bed. Then the optical power of whole system,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Now, since this system serves as telescope, the optical power of the system must be equal to zero.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Since the diam eter D of the ob jective is 2R1 and thatof the eye-piece is d = 2R2 So, the magnification,
Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.61. Two identical thick symmetrical biconvex lenses are put close together. The thickness of each lens equals the curvature radius of its surfaces, i.e. d = R = 3.0 cm. Find the optical power of this system in air. 

Ans. Optical powers of the two surfaces of the lens are

 

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

and optical power of the combination of these two thick lenses,

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

So, power of this system in air is, Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

Q.62. A ray of light propagating in an isotropic medium with refractive index n varying gradually from point to point has a curvature radius ρ determined by the formula 

 

 Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

 

 where the derivative is taken with respect to the principal normal to the ray. Derive this formula, assuming that in such a medium the law of refraction n sin θ = const holds. Here θ is the angle between the ray and the direction of the vector Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev at a given point. 

Ans.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev
 

Q.63. Find the curvature radius of a ray of light propagating in a horizontal direction close to the Earth's surface where the gradientof the refractive index in air is equal to approximately 3.10-8  m-1. At what value of that gradient would the ray of light propagate all the way round the Earth? 

Ans.

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

Irodov Solutions: Photometry and Geometrical Optics- 3 Notes | EduRev

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