# Reflection of Light Short Trick | 7tricks

Today we will tell you how to solve the question related to the reflection of light easily!

So, we will tell you step-by-step in an easy way to solve that quickly.

We divided that into some steps!

Step

First, to solve the question of light we have to knowledge about u, v, f and m.

So, here is some definition of these,

v = distance of the image from the mirror It can be positive and negative It can be decided according to the coordinate system.

u = It is the distance of the object from the mirror.

Golden rule:- It is also negative if in question, we have to find u then we take u always negative.

f = It is the length between the Focus and pole. The focal length of a spherical mirror is equal to half of its radius of curvature. m = It denotes magnification. The magnification of a spherical mirror is the ratio of the height of the image to the height of the object

Step

We have to know that in the concave mirror Focus is in LHS so we take the focal length as negative but in a convex mirror, Focus is in RHS so we take the focal length as positive.

[Trick:- Try to learn it by this  – ‘In you left there is a cave of loin’ Cave means concave and it is in left so other will be in right.]

(These RHS & LHS are taken with a coordinate system, In +ve x-axis and +ve y-axis all quantities are positive &  in the -ve x-axis and y-axis all quantities are negative.)

Step

We have use this formula-

1/v + 1/u = 1/ f

This mirror formula gives the relation between the object-distance, image-distance, and focal-length

Step 4 :-

For finding magnification, Use

m = -v/ u

here, v=Height of image & u=Height of object

#### SOME  EXAMPLES –

Que. A convex mirror (used for rear-view) on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00m from this mirror, find the position, nature and size of the image.

Sol.

Given,

Radius R = 3.00 m Object distance, u = -5.00 m

Find that, Image distance v = ? Height of image h ‘ = ?

We know that, Focal length, f = R/2 = 3.00/2 = 1.50 M

Since, 1/v + 1/u = 1/f

1/v = 1/f – 1/u

= 1/ 1.50 – 1/ (-5)

= 1.50 = 1/ 5

= 5 + 1.50/7.50

v = 7.50/6.50 = 1.15

The image is 1.15 m at the back of the mirror magnification, m = -v/u = 0.23

The image is virtual, erect and smaller in size by a factor of 0.23.

Hope you understand it correctly!