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REFLECTION OF LIGHT
PART-3
CLASS-10TH
BY R K CHAUDHARI
R K CHAUDHARY
rkchaudhary222@gmail.com
Contact no:8960502268
The Mirror Equation: Relation between u, v, f
Mirror Equation
1/f = 1/v +1/u
Where u- Distance from Pole to the object
v- Distance from pole to the image
f- Distance from pole to the focus
New cartesian sign convertion
1-Mirror are always left side object position
2- All distance measure poles
3- Any Ray left to right then magnitude +ve
4- Right to left side magnitude -ve
5- Principal axis always above height always +ve
6- Principal axis below height is -ve
R K CHAUDHARY
rkchaudhary222@gmail.com
Contact no:8960502268
C P
LEFT SIDE
Right Side
From the above convention, we can deduce that:
 u is +ve if object is real
 u is -ve if object is virtual,
 v is +ve if image is real
 v is -ve if image is virtual
 f and R are +ve if focus is real (C is in front of the mirror like in a
concave mirror).
 f and R are -ve if focus is virtual (C is behind the mirror like in a
convex mirror).
Q-1 An object of height 4cm is placed at 30 cm from a concave
mirror of focal length 20 cm. Find the position of the image.
Solve- u = +30 cm
v = ?
f = +20cm
mirror formula
1/f = 1/v + 1/u
1/20 = 1/v +1/30
1/v = 1/20 -1/30
v = 600/10
v = +60cm
therefore, image is formed at a distance of +60cm in front of
the mirror.
R K CHAUDHARY
rkchaudhary222@gmail.com
Contact no:8960502268
Magnification- it is ratio of the height of image to the height of
object. It is denoted by m
m = height of image/ height of object
m =h’/h = - v/u
m = - v/u
m can be either positive or negative, depending on the nature of the
image. If m is positive, h’ and h have the same sign. This means that
the image is formed on the same side of the principal axis as the
object. Or the image is erect.
Q- Above example
m = h’/h = -v/u
h’= -(v/u)*h
= -(+60/+30)*4
= -8 cm
Height of the image is 8 cm and it is inverted. As v is +ve , the image
is real
R K CHAUDHARY
rkchaudhary222@gmail.com
Contact no:8960502268
Practice question
1-A point source of light is kept in front of a convex mirror of
radius of curvature 40 cm. The image is formed at 10 cm behind the
mirror. Calculate the object distance.
2-Ram and sham are standing in front of a mirror. Ram is at a
distance 4m from the mirror. Distance between sham and image of
Ram is 10m. What is the distance between Ram and sham?
3- An object is placed 0.28m from a concave mirror whose focal
length is 0.1m. Find where the image is formed? Is it real or virtual?
4-Draw the ray diagrams for all possible positions of the object
in case of concave mirror.
5-State mirror formula. How does ‘f’ change when object
distance ‘u’ from the mirror is changed?
6-Find the position, size and the nature of the image formed by
a spherical mirror from the following data.
u = -20 cm f = -15 cm ho = 1.0 cm
R K CHAUDHARY
rkchaudhary222@gmail.com
Contact no:8960502268

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Reflection of light part-3

  • 1. REFLECTION OF LIGHT PART-3 CLASS-10TH BY R K CHAUDHARI R K CHAUDHARY rkchaudhary222@gmail.com Contact no:8960502268
  • 2. The Mirror Equation: Relation between u, v, f Mirror Equation 1/f = 1/v +1/u Where u- Distance from Pole to the object v- Distance from pole to the image f- Distance from pole to the focus New cartesian sign convertion 1-Mirror are always left side object position 2- All distance measure poles 3- Any Ray left to right then magnitude +ve 4- Right to left side magnitude -ve 5- Principal axis always above height always +ve 6- Principal axis below height is -ve R K CHAUDHARY rkchaudhary222@gmail.com Contact no:8960502268 C P LEFT SIDE Right Side
  • 3. From the above convention, we can deduce that:  u is +ve if object is real  u is -ve if object is virtual,  v is +ve if image is real  v is -ve if image is virtual  f and R are +ve if focus is real (C is in front of the mirror like in a concave mirror).  f and R are -ve if focus is virtual (C is behind the mirror like in a convex mirror). Q-1 An object of height 4cm is placed at 30 cm from a concave mirror of focal length 20 cm. Find the position of the image. Solve- u = +30 cm v = ? f = +20cm mirror formula 1/f = 1/v + 1/u 1/20 = 1/v +1/30 1/v = 1/20 -1/30 v = 600/10 v = +60cm therefore, image is formed at a distance of +60cm in front of the mirror. R K CHAUDHARY rkchaudhary222@gmail.com Contact no:8960502268
  • 4. Magnification- it is ratio of the height of image to the height of object. It is denoted by m m = height of image/ height of object m =h’/h = - v/u m = - v/u m can be either positive or negative, depending on the nature of the image. If m is positive, h’ and h have the same sign. This means that the image is formed on the same side of the principal axis as the object. Or the image is erect. Q- Above example m = h’/h = -v/u h’= -(v/u)*h = -(+60/+30)*4 = -8 cm Height of the image is 8 cm and it is inverted. As v is +ve , the image is real R K CHAUDHARY rkchaudhary222@gmail.com Contact no:8960502268
  • 5. Practice question 1-A point source of light is kept in front of a convex mirror of radius of curvature 40 cm. The image is formed at 10 cm behind the mirror. Calculate the object distance. 2-Ram and sham are standing in front of a mirror. Ram is at a distance 4m from the mirror. Distance between sham and image of Ram is 10m. What is the distance between Ram and sham? 3- An object is placed 0.28m from a concave mirror whose focal length is 0.1m. Find where the image is formed? Is it real or virtual? 4-Draw the ray diagrams for all possible positions of the object in case of concave mirror. 5-State mirror formula. How does ‘f’ change when object distance ‘u’ from the mirror is changed? 6-Find the position, size and the nature of the image formed by a spherical mirror from the following data. u = -20 cm f = -15 cm ho = 1.0 cm R K CHAUDHARY rkchaudhary222@gmail.com Contact no:8960502268