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Physics

Hochschule
Sixth Year / 12th Grade

Ray optics full notes which will make ur doubts clear. It is easy and simple. It will best best for boards and competitive exams.

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PHYSICS NOTES www.gneet.com

RAY OPTICS AND OPTICAL INSTRUMENTS
SECTION I
REFLECTION OF LIGHT
Nature of light
- Light is an electromagnetic radiation which causes sensation in eyes.
- Wavelength of visible light is 400 nm to 750 nm
- Speed of light in vacuum is highest speed attainable in nature 3.0 ×108 m/s
- Wavelength of light is very small compared to the size of the ordinary objects, thus
light wave is considered to travel from one point to another along straight line
joining two points.
- The straight path joining two points is called ray of light.
- Bundle of rays is called a beam of light
- Light show optical phenomenon such as reflection , refraction, interference and
diffraction
REFLECTION OF LIGHT BY SPHERICAL MIRRORS
Law of reflection. I) The angle of incidence (angle between incident ray and normal to the
surface) and angle of reflection (angle between reflected ray and normal to the surface)
are equal.
ii) Incident ray, reflected ray and normal to the reflecting surface at the point of incidence
lie in the same plane.
Note : Normal to the curved surface always passes through the centre of curvature
Sign convention: (i) All distances are measured from the pole of the mirror.
(ii) The distance in the direction of incidence light is taken as positive.
(iii) Distance in opposite to the direction of incident light is taken as negative
(iv) Distance above the principal axis is taken as positive.
(v) Distance below the principal axis taken as negative.
FOCAL LENGTH OF SPHERICAL MIRROR
Let P the pole of concave mirror, F be focal point
and C be the centre of curvature.
Consider incident light parallel to principal axis
strikes the mirror at point M and reflected rays
passes through focal point F.
MD is perpendicular from M on principal axis.
Let ∠MCP = θ
Then from geometry of figure, ∠MFP = 2θ.
MD MD
Now tan θ= and tan2 θ = –eq(1)
CD FD
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For small θ , tan θ = θ and tan2 θ = 2 θ

MD MD
Therefore from eq(1) 2 =
CD FD
Thus FD = CD/2 But CD = R and FD = f
Thus f = R/2

The Mirror equation
As shown in figure AB is object while
A’B’ is image of the object . AM and AN
are two incident rays emitted from
point A. Let F be focal point and FD = f
focal length
For small aperture , DP will be very
small neglecting DP we will take FD =
FP = f
I) In ∆ A’B’F and ∆MDF are similar as
∠MDF = ∠ A’B’F
𝐴′𝐵′ 𝐵′𝐹
=
𝑀𝐷 𝐹𝐷
As MD = AB and FD = FP
𝐴′𝐵′ 𝐵′𝐹
=
𝐴𝐵 𝐹𝑃
ii) In ∆ A’B’P and ∆ ABP are also similar. Therefore
𝐴′𝐵′ 𝑃𝐵′
=
𝐴𝐵 𝑃𝐵
Thus comparing above two equations
𝐵′𝐹 𝑃𝐵′
=
𝐹𝑃 𝑃𝐵
But B’F = PB’-PF
𝑃𝐵′ − 𝑃𝐹 𝑃𝐵′
= − 𝑒𝑞(1)
𝐹𝑃 𝑃𝐵
From sign convention
Focal length = PF = -f
Image distance PB’ = -v
Object distance = PB = -u
Putting the values in equation (1)
−𝑣 − (−𝑓) −𝑣
=
−𝑓 −𝑢
𝑣−𝑓 𝑣
=
𝑓 𝑢

2
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1 1 1
+ =
𝑣 𝑢 𝑓
This relation is known as mirror equation

Magnification
In ∆ A’B’P and ∆ ABP are also similar. Therefore
𝐴′𝐵′ 𝑃𝐵′
=
𝐴𝐵 𝑃𝐵
From sign convention
PB’ = image distance =-v
PB = object distance = -u
A’B’ = size of image = -h ‘
AB = size of object = h
−ℎ′ −𝑣
=
ℎ −𝑢

ℎ′ 𝑣
=−
ℎ 𝑢

Magnification, m = size of the image / size of object = h’/h
𝑣
𝑚=−
𝑢
Note : If m is negative , image is real and inverted
If m is positive image is virtual and inverted
If |m| = 1 , size of the object = size of image
If |m| >1 size of image > size of the object
If |m| < 1 size of the image < size of the object

Solved Numerical
Q) An object is placed in front of concave mirror at a distance of 7.5 cm from it. If the real
image is formed at a distance of 30 cm from the mirror, find the focal length of the mirror.
What should be the focal length if the image is virtual?
Solution: Case I: When the image is real
U = -7.5 cm; v=-30cm; f= ?
We know that
1 1 1
= +
𝑓 𝑣 𝑢
1 1 1
= +
𝑓 −30 −7.5

3
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1 −5
=
𝑓 30
f = -6 cm
The negative sign shows that the spherical mirror is convergent or concave
Case II : When image is virtual
u = -7.5 cm ; v = +30 cm
We know
1 1 1
= +
𝑓 𝑣 𝑢
1 1 1
= +
𝑓 30 −7.5

1 −3 −1
= =
𝑓 30 10
f = -10 cm
Q) An object 0.5 cm high is placed 30 cm from convex mirror whose focal length is 20 cm.
Find the position, size and nature of the image.
Solution: We have
U = -30 cm, f = +20 cm
Form mirror formula
1 1 1
= +
20 𝑣 −30
1 1
=
𝑣 12
V = 12 cm
The image is formed 12 cm behind the mirror. It is virtual and erect
m = h’/h = -v/u = -12/30
ℎ′ −𝑣 −12
𝑚= = = = 0.4
ℎ 𝑢 −30
h’ = mh = 0.4 ×0.5 = 0.2 cm
positive sign of m indicate image is erect.

Q) A thin rod AB of length 10 cm is placed on the principal axis of a concave mirror such
that its end B is at a distance of 40 cm from the mirror and end A is further away from the
mirror. If the focal length of the mirror is 20cm, find the length of the image of rod
Solution: given f = -20 cm, distance of B = u1 =- 40 cm,
Since B is at centre of curvature image will be formed at -40 cm
Distance of A u2 = -50 cm
1 1 1
= +
−20 𝑣 −50
V = -33.3 cm
Image of A is also on the side of object , Now length of image = 40-33.3 = 6.70 cm
4
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