PHYSICS 10TH CH10 LIGHT - TOPIC 1.10 - 1.12
THE PHYSICS CAFE
CHAPTER TEN - LIGHT - REFLECTION AND REFRACTION
1.10 Sign Convention for Spherical Mirrors**
We will have to use some sign convention for solving the mirrors related problems is called sign convention.
Assume origin of Cartesian coordinate
system is concentric with pole of mirror and the object is placed at left side
of mirror so that light ray fall on mirror from left-hand side. Then following
conventions are
- All the distances are measured from pole.
- All the distance measured right sides of the pole are taken as positive.
- All the distance measured left sides of the pole are taken as negative.
- All the distance measured above the pole (principal axis) is taken as positive.
- All the distance measured below the pole (principal axis) is taken as negative.
Hint: If we move from
pole/origin, along positive axis, distances are positive and along negative
axis, distances are negative.
1.11 Mirror Formula and Magnification (Derivation not required) **
Distance of object from the pole is
called Position of object (u)
Distance of image from the pole is
called Position of image (v)
Perpendicular distance of object’s head
from the pole/principal axis is called Height of object (ho)
Perpendicular distance of image’s head from the pole/principal axis is called Height of image (hi)
The relation between position of
object, position of image and focal length of mirror is called Mirror formula
Ratio of height of image to height of
object is called Magnification of mirror
It tells that our mirror made image how
many times bigger or smaller of the given object. If magnification of a given 2
it means image will be twice taller of object.
Height of image (hi) Position
of image (v)
Magnification = ___________________ = - ___________________
Height of Object (ho) Position of object (u)
hi v
m = ___ = - ____
ho u
Hint: Magnification can be Negative or Positive both
For real image: u is –ve and v is also –ve by putting value in above formula we have, m = -ve or ho is +ve and hi is –ve by putting value in above formula we have, m = -ve ho
For virtual image: u is –ve and v is +ve by putting value in above formula we have, m = +ve or ho is +ve and hi is also +ve by putting value in above formula we have, m = +ve
1.12 Practice Problems on Image Formation in Concave and Convex Mirror**
To
see how to solve practice problems in video lectures click here…
Hint: before solving any problem please apply sign convention for u, v, ho, hi, f and m etc to get a correct result.
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