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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.
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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 –
- Specular or Regular Reflection – In this light ray strike a smooth surface or very shiny surface then reflected rays are parallel to each other.
- 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:
- Incident, reflected ray & normal lies in the same plane.
- 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
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