Light-Reflection and Refraction Important Questions

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Light is a form of energy and it is responsible for our sense of vision. Main source of light on the earth is the sun. Light always travels in a straight line. Light can pass easily through transparent objects, partially through translucent objects and cannot pass through opaque objects. 

Read Also: Lens Formula and Magnification


Reflection of Light

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When light rays strike a surface and bounce back in the same medium. The reflection of light is of types –

  1. Specular or Regular Reflection – In this light ray strike a smooth surface or very shiny surface then reflected rays are parallel to each other.
  2. Irregular Reflection – In this light ray strike on irregular surfaces then the reflection rays are non – parallels to each other.

Law of Reflection:

Angle of Incident is equal to angle of reflection that is Li = Lr. Incident rays, reflected rays and normal rays lie in the same plane.

Refraction of Light:

When the light rays travel from one medium to another then it bends towards the normal or away from the normal this phenomenon is called refraction of light.

Laws of Refraction:

  1. Incident, reflected ray & normal lies in the same plane.
  2. The sine value of the angle of incidence and that of angle of refraction is always constant. The ratio is called the Refraction Index and the law is also called Snell's Law.

Explanation:

Incident Ray – Light rays which fall on a reflecting surface are called incident rays.

Point of Incident – The point on the reflecting surface at which the incident ray falls.

Angle of Incident – Angle between incident ray and normal ray is known as angle of incident and it is represented by /_ i.

Read More About Image Formation by Lenses


Important Sample Questions

Question. Focal length of concave lens is 15 cm at what distance the object should be placed so that image is 10cm. Find the nature of the image. (2 marks)

Sol: F = -15cm

U =?

V = -10cm

1/F = 1/v – 1/u

-1/15 = -1/10 – 1/u

-1/15 + 1/10 = - 1/u

-2+3/30 = - 1/u

1/30 = - 1/u

-30m =u 

Q2. The Refractive Index of glass with respect to air is 3/2. Find out the refractive index of air with respect to glass. (2 marks)

Sol: aNg = 3/2

gNa =?

aNg * gNa = 1

3/2 * gNa =1

gNa = 2/3 m/sec

Q3. A 2cm tall object is placed perpendicular on the principal axis of a convex lens of focal length of 10cm. The distance of the object from lens is 15cm. Find out nature, position and size of image and magnification? (3 marks)

Sol: O = 2cm

F = 10cm

U = - 15cm

1/f = 1/v – 1/u

1/10 = 1/v – (-1/15)

1/10 = 1/v + 1/15

1/10 – 1/15 = 1/v

3 – 2/30 = 1/v

1/30 = 1/v

V = 30cm 

Image is real and at 30cm from lens.

I/O O/I = v/u

I/2 = -30/15

I*30 = -15*2

I = -15 * 2/30

= -4m

The size of the image is 4cm and it's inverted.

M = v/u 

m = - 30/15

= 2 times real in magnification.

Q4. A 50cm tall object is placed on the principal axis of a concave lens it 20cm tall image form on the screen placed at a distance 10cm from the lens calculate the focal length of the lens? (2 marks)

Sol: O = 50cm

I = -20cm

V = 10cm

F =?

I/O = v/u

-20/50 = 10/u

-20 * u = 50 * 10

U = 50* 10 /20

U = - 25cm

I/f = 1/v – 1/u

1/f = 1/10 – (-1/25)

1/f = 1/10 + 1/25

1/f = 5 + 2/50

1/f = 7/50

F = 50/7

F = 7.14cm Answer.

Q5. Size of the object is 2cm and it is placed on the principal axis of the convex lens of focal length 10cm. Find out the natural position of an object placed at 15cm. (2 marks)

Sol: u = -15cm

f= 10cm

O= 2cm

1/f = 1/v – 1/u

1/10 = 1/v – (-1/ -15)

1/10 + 1/10 = 1/v

2+3/30 =1/v

1/6 = 1/v

m = v/u = I/O

m = -6/15 = I/2

m = 6*2/-15 = I

m = -12/15

-12/15 = I

Q6. Refractive index of diamond is 2.42 explain the meaning of this statement? (2 marks)
Sol: aNd = 2.42

aNd= Speed of light in air

Speed of light in diamond

aNd = 2.42 – Speed of light in diamond

3 * 108 /242 m/s 

aNd = 300/242 * 108

Speed of light in diamond is less than the speed of light in air.

Q7. A student uses lenses of focal length of 40cm to find out the power of lens? (1 mark)

Sol: Focal length = 40cm (Convex lens)

Power of lens = 100/ focal length

= 100/40

= 2.5 D

Q8. A ray light enters into beryllium from air its R.I is 150. Find out by what percent speed of light in beryllium is less than what it is in the air? (2 marks)

Sol: anb = 1.50

anb = Speed of light in air 

Speed of light in beryllium

b = 30*10 8/1.50

b = 2*108 m

less = 3*108 – 2*108

less % = less* 100/ original

= 1*10 8 *100/ 3 *108

Read More:

Convex Mirror Refraction of Light
The Refractive Index Reflection of Light

CBSE X Related Questions

  • 1.

    Select TRUE statements about lymph from the following: 
    A. Lymph vessels carry lymph through the body and finally open into larger arteries. 
    B. Lymph contains some amount of plasma, proteins and blood cells. 
    C. Lymph contains some amount of plasma, proteins and red blood cells. 
    D. Lymph vessels carry lymph through the body and finally open into larger veins. 
    The true statements are:

      • A and B
      • B and D
      • A and C
      • C and D

    • 2.
      Name the type of lenses required by the persons for the correction of their defect of vision called presbyopia. Write the structure of the lenses commonly used for the correction of this defect giving reason for such designs.


        • 3.
          Show the formation of magnesium chloride by electron transfer. Write the name of the cation and anion present in the compound formed. (Atomic Number of Mg = 12, Cl = 17)


            • 4.
              State the reason for the bending of the refracted ray away from the normal.
              path of a ray of light going from Medium 1 to Medium  2


                • 5.

                  Express refractive index of Medium 2 with respect to Medium 1 in terms of speed of light in two media. 


                    • 6.

                      Give reasons:  
                      (i) The sky appears dark to passengers flying at very high altitudes. 
                      At very high altitudes, passengers are above the atmosphere where there is less scattering of sunlight. As a result, they do not see the scattered blue light and the sky appears dark, similar to the condition experienced by astronauts in space. 
                      (ii) 'Danger' signal lights are red in color. 

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