More Challenging Problems on Reflection and Refraction

Introducing Optics: Part 3 (More Challenging Problems on Reflection and Refraction)

General Physics, High School and College Freshman level

Target Audience: These notes might be of use to Indian students preparing for the ISC or CBSE Class 11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students from across the world preparing for their A Level Examinations, IB (International Baccalaureate) or AP Physics.

This might also be helpful in studying topics required by Common Core Physics. 

More Challenging Problems on Reflection and Refraction: Notes, Figures and Problems with Solutions

Target Audience: These notes on Atomic Structure are meant for college freshmen, or high school students in Grades 11 or 12.

They might be of use to Indian students preparing for the ISC or CBSE Class 11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students from across the world preparing for their A Level Examinations, IB (International Baccalaureate) or AP Physics.

This compilation of notes has been prepared by Anirud Thyagarajan of IIT Kharagpur

Refraction and Reflection Revisited: A few Challenging Problems

  1. Determine the focal length of a concave mirror if with the distance between an object and its image being equal to l = 15 cm, the transverse magnification Ω = -2.0.


From the mirror formula, we have

1/s + 1/s’ = 1/f   we get f = ss’/(s+s’)    _____(1)

In accordance with the given problem, s-s’ = l*s’/s = Ω .

From these two relations,

S = 1/(l - Ω) and s’ = -l Ω /(1+ Ω).

Putting it back in eqn (1), we get

F = l Ω/(1- Ω)^2  = -10 cm.

  1. Determine the focal length of a concave mirror if in a certain position of the object the transverse magnification is β= --0.50 and in another position displaced with respect to the former by a distance l = -5.0 cm the transverse magnification µ = -0.25.


According to the mirror formula, we obtain:

1/s’ + 1/s = 1/f .               so, s/s’ + 1 = s/f.

So, 1/ β = s/f – 1 = (s – f)/f

  • β = f/(s – f)

Now, it is evident from the above form, that for smaller ,s must be large, so that the object is displaced away from the mirror in second position.


µ = f/(l + s – f)

Thus, eliminating s from the two equations, we get

F = l β µ/(β - µ) = -2.5 cm.

  1. Illustration

: A plane mirror is placed 22.5 cm in front of a concave mirror of focal length 10 cm. Find where an object can be placed between the two mirrors, so that the first image in both the mirrors coincides.





As shown in figure, if the object is placed at a distance x from the concave mirror, its distance from the place mirror will be (212.5 - x). So, plane mirror will form equal and erect image of object at a distance (22.5 - x) behind the mirror.


Now as according to given problem the image formed by concave mirror coincides with the image formed by concave mirror, therefore for concave mirror


      v = - [22.5 + (22.5 - x)] = - (45 - x)    and    u = - x



i.e. x


- 45x + 450 = 0    or    (x - 30)(x - 15) = 0


i.e. x = 30 cm      or      x = 15 cm


But as the distance between two mirrors is 22.5 cm, x = 30 cm is not admissible. So the object must be at a distance of 15 cm from concave mirror.

  1.  A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes 9 m/s vertically above it. If the refractive index of water is 4/3 find the actual velocity of the dive of the bird.



Solution: If at any instant the fish is at the depth 'x' below fish water surface while the bird at a height y above the surface, then the apparent height of the bird from the surface as seen by the fish will be given 
µ =       or
So, the total apparent distance of the bird as seen by the fish in water will 
or       or
  • 9=3+µ()

  1. Illustration

: A ray of light is incident on one face of a prism (µ = 1.5) at an angle of 600

. The refracting angle of the prism is also 600

. Find the angle of emergence and the angle of deviation. Is there any other angle of incidence which will produce the same deviation ?





Angle of incidence = i = 600

at point P,

=> sin r1=


or  r1 350

6'      using     r1+ r2= A


At point Q,    

=> sin e = 1.5 sin 240

44 sin e = 0.63


Deviation == (i + e) - A = (60 + 39) - (60) = 390

If i and e are interchanged, deviation remains the same. These same deviation is obtained for angles of incidence 600and 390


  1. Calculate the dispersive power of crown and flint glass-prism from the following data: for crown glass                        for crown glass 

    µV = 1.522; µR = 1.514                                   µV = 1.662; µR = 1.644 
    Solution: For crown glass      µV = 1.522; µR = 1.514 

     µy =  = 1.518 

    Hence, the dispersive power of crown glass

    W =  = 0.01544 

    for flint glass    µV = 1.662; µR = 1.644 

     µ =  = 1.6 

  1. Illustration: Find the angle of a prism of dispersive power 0.021 and refractive index 1.53 to form on achromatic combination with the prism of angle 4.20 and dispersive power 0.045 having refractive index 1.65. Also calculate the resultant deviation. 
    Solution: W = 0.021; µ = 1.53     W' = 0.045; µ' = 1.65 

    A' = 4.2 0

    For no dispersion 

    W + W'' = 0 

    or W(µ - 1)A + W'(µ' - 1)A' = 0 

    or A = -  = - 11.40 

    Net deviation =  + ' = (µ - 1)A + (µ' - 1)A' 

    = - 11.040(1.53 - 1) + 4.20(1.65 - 1) = - 3.120

  1. Illustration: A transparent rod 40 cm long is cut at one end and rounded to a hemispherical surface 12 cm radius at the other end. A small object is embedded within the rod along its axis and half way between its ends. When viewed from the end of the rod, the object appears 12.5 cm deep. What is its apparent depth when the object viewed from curved end ?
    Solution: Case - I: When the object viewed from the flat surface:

Real depth of the object = 20 cm 

Apparent depth = 12.5 cm 

Using µ =     we have µ =  = 1.6 
Case - II: When the object is viewed the curved surface: 

Here the refraction is taking place at the single curved surface. So we will use 


Here µ1 = 1.6; µ2 = 1; u = -2 cm; v = ? ; R = - 12 cm 


=> v = - 33.3 cm 

Hence the object appears 33.3 cm deep from the curved surface. 

  1. Illustration: A thin hollow equiconvex lens silvered at the back converges a parallel beam of light, at a distance of 0.2m in front of it. Where will it converge the same light, if filled with water (µ = 4/3). 

    Solution: When lens is hollow, it will simply act as concave mirror

 Parallel beam is focused at focus of mirror only 
 f=-0.2m( R=2f=-0.4m) 
When filled with water 
Pnet =2PL +PM =2  
Now the complete system acts as mirror 
 fnet =  =-0.12m 
 Parallel beam will now be focused at a distance of 12 cm from the lens (on the side of object). 

  1. Two plane concave lens of glass (µ = 3/2) have radius of curvature 20 & 30 cm, they are placed in contact with corved surface towards each other and the space between them is filled with a liquid of refractive index 4/3. Find the focal length of the system. 


    Solution:   Pnet = P1 + P2 + P3 
    fnet =  = - 72 cm    (focal length of system) 

    11.   A small fish 0.4 m below the surface of a leke is viewed through a simple converging lens of f = 3 m. The lens is kept at 0.2 m above the water surface such that the fish lie on the optical axis of the lens. Find the image of the fish seen by observer, the refractive index of water = 4/3.

    Solution: Apparent depth of fish =
    = 0.3

    Apparent position of object = - (0.3 + 0.2) = - 0.5

    image of the fish is formed at same point where fish is present.

  1. A concave lens focusses a distant object 40 cm from it on a screen placed 10 cm away from it. A glass plate (µ = 1.5) and of thickness 3 cm is inserted between lens and a screen, where should the object be placed, so that its image is again focussed on the screen.



    f = 8 cm

    Shift due to glass slab = t

    = 1 cm
    For this position of object, image would have formed at 10 - 1 = 9 cm
    if glass slab was not present

    u = - 72 cm
    Object should be placed 72 cm from lens.

  1. A spherical convex surface separates object and image space of refractive index 1.0 and 4/3. If radius of curvature of the surface is 10 cm, find its power.

    Solution: For finding focus, we choose our object to be at

    , then image is formed at focus




    P =

    P = 2.5 dioptre

  1. An isotropic point source is placed at a depth h below the water surface floating opaque disc is placed on the surface of water so that the bulb is not visible from the surface. What is the minimum radius of the disc. Take refractive index of water = µ. 


    Light from the source will not emerge out of water if I > C 
    Therefore, minimum radius R correspond to I = C 
    In SAB,
     = tanC 
    R = h tanC               

  1. The x - y plane is boundary between two transparent media. Medium-1 with z  0 has a refractive index  and medium 2 with z  0 has refractive index . A ray of light inmedium-1 given by vector  is incident on the plane of separation, find the unit vector in the direction of the refracted ray in medium-2. 
    Solution: Let refracted ray br  

    Normal to plane of incident and normal =  


    it must also be normal to refracted ray 


     cos ( - i) =  

     = cos1200s

     i = 600 

    r = 450 

    Now since angle between refracted ray and Normal = 450 

     c2 = a2 + b2 = a2 +  


      c =  

  1. A ray of light in air is incident atgrazing angle (i = 900) on a long rectangular slab of a transparent medium of thickness t = 1.0 m. The point of incidence is the origin   A(0, 0). The medium has a variable index of refraction n(y) given by n(y) =  where k = 1.0 m-3/2. The refractive index of air is 1. (i) Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the point. (ii) Obtain an equation for trajectory y(x) of the ray in the point. (iii) Determine the co-ordinates (x, y1) of the point P where the ray intensects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently. 


    Taking on arbitrary point P(x, y) refractive index at this point n = 

    from Snell's law     n sin = constant      applying this for initail pt. (when ray is entering medium B) and at point. 

    1 x sin900 =  sini 

     sin i =       it can be seen that i =  

    s  Slope = tan = cot i = 
    (ii)  = cot i =  


    it passes through origin        C = 0 

      x = 4    is the equation of trajectory 

    when ray comes out of the mediums

    then    x = 4 x 1 = 4 

      Co-ordinate of pt- is (4, 1) 

    If medium on both sides are same, then angle with which the ray enters the medium = angle with which the ray comes out. 

      Ray will be parallel to x-axis.