class 10 - Numericals on Light Reflection and Refraction

Numerical Problem on Light Reflection and Refraction

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Q1: How fast does the light travel in a glass of refractive index 1.5?

Answer:
  By Snell's law, refractive index (n) is the ratio of i.e.

 






Q2:When a bright object is placed 10 cm away from a concave mirror, its real image is formed at a distance 40cm from the mirror. What is the focal length of the mirror?

Answer:
     







Q3: If the angle of incidence of a ray of light falling on the glass surface is 30° and the angle of refraction is 19°. what is the refractive index of glass?

Ans:
      angle of incidence  i = 30°
      angle of refraction  r = 19°
      refractive index   n = sin i/sin r  = sin 30/sin 19
                                  n = .5/.3256 = 1.535               ....(ans)

Q4: If a child crawls towards a mirror at a rate of 0.20 m/s, then at what speed will the child and her image come nearby to each other?

Ans: In 1 sec, the child moves towards the mirror by a distance of 0.20 meters. In the same second, her image also moves 0.20m closer.
Therefore, the child and her image comes closer to each other by 0.40m per second.



Q5: How many images will you see when two plane mirrors are perpendicular to each other?
Ans:
       The answer is three. You may see practically as shown in the figure. You may follow the link on multiple reflections to see more such images. 

For two mirrors placed at any angle, the number of images formed by the mirrors can be determined by:

 N = (360°/θ ) - 1

where θ is the angle between the mirrors and N is the number of images formed. For parallel mirrors θ = 0, hence N = ∞. (You may have seen this effect in barber shops).

Q6: A converging lens has a focal length of 15cm. An object is placed 60cm from the lens. Determine the image.

Ans:       Object Distance is u = -60cm
               focal length f = 15 cm

Lens formula is:  1/v - 1/u = 1/f
                1/v = 1/15 + (-1/60)
                      = 1/20
 ⇒            v = 20 cm
The image is real (+ve v) and is formed on the other side of the lens.
The magnification is:  m = 20/-60 = -1/3
The -ve sign of m tells that the image is inverted. The magnitude of m (magnification) is less than one, thus image formed is of reduced size.

Q7: What is the focal length of a lens that produces a real image three times as large as the object if the distance between image and object is 1.0m?

Ans:  magnification m = 3
          Let u be the object distance, the image distance will be v = 1-u
          Since image is real, we can assume it is a convex lens.
           m = 3 = v/u = (1-u)/u
    ⇒    u = -025cm and  v = 0.75cm
         1/v - 1/u = 1/f
     ⇒  f = 18.75cm

Q8(*): At what value of the angle of incident α1 is a beam of light reflected from the surface of water perpendicular to the refracted beam? Assume n is the refractive index of the refracting medium.

Answer:
      Let us try to draw a ray diagram to understand the problem.

Let XY be the plane which separates the two media. The incident ray AO gets reflected partly as OB and gets refracted as ray OC. NOM is normal to the plane.

incident angle  ∠AON = α1
reflected angle ∠NOB = α2
refracted angle ∠COM = α3

Since ∠COB = π/2               ...(given)
α1 = α2                                 ...(2nd law of reflection)
This gives α2 = π/2 - α1

According to Snell's law (2nd law of refraction),
n = sin i/sin r = sin α1/sin α2
⇒ n = sin α1/sin ( π/2 - α1)}
⇒ n = sin α1/ cos α1
$ n = tan α1
α1 = tan^-1 n

Q 9: Using mirror formula, compute the position of the object placed in front of a concave mirror of focal length f so that the image formed is of same size of the object.

Answer: In this case, magnification m = 1, since image size = object size (given).

Q10: Water has refractive index = 1.33 and air has refractive index = 1.00. Find the critical angle for a water-air boundary.

Answer: Critical angle is considered when a ray of light travels from denser medium to refractive medium.
See the figure,

the angle of incidence ∠i = θ_c and ∠r = 90°
Applying snell's law ⇒ μ₁ sin θ_c = μ₂ sin 90°
⇒ 1.33 × sin θ_c = 1.00 × sin 90°
⇒ sin^-1 θ_c = 0.752
⇒ θ_c = 48.7°

Q11 (NCERT): Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:  Given, radius of curvature = 32 cm
 Focal Length = radius of curvature / 2 = 32 /2 = 16 cm


Q12(NCERT): A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Answer: Given mirror = concave mirror
Distance of the object from mirror = - u = - 10 cm
Distance of the image from the concave mirror = v?

Magnification (m) = -3

m = -v/u
⇒ -3 = -v/-10 cm
⇒ v = -3 × 10 = -30 cm
Thus image is formed 30 cm from pole in front of the concave mirror (-ve sign tells that image formed is on the same side of the object).


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