Question

List the new cartesian sign convention for reflection of light by spherical mirrors. Draw
a diagram and apply these conventions for calculating the focal length and nature of a
spherical mirror which forms a ' times magnified virtual image of an object placed
18 cm in front of it.​

Answer 1

Explanation:

The following sign convention is used for measuring various distances in the ray diagrams of spherical mirrors:

Object is always placed to the left of mirror

All distances are measured from the pole of the mirror.

Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative.

Distances measured along y-axis above the principal axis are positive and that measured along y-axis below the principal axis are negative.

Since the image formed is magnified and virtual, the mirror is convex.

According to the convention

u=−18cm

m=1/3=−

u

v

⟹v=6cm

From mirror equation,

v

1

+

u

1

=

f

1

⟹f=

u+v

uv

⟹f=9cm

solution

Answer 2

Answer:

the following sign convention is used for measuring various distances in the ray diagrams of spherical mirrors:

Object is always placed to the left of mirror

All distances are measured from the pole of the mirror.

Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative.

Distances measured along y-axis above the principal axis are positive and that measured along y-axis below the principal axis are negative.

Since the image formed is magnified and virtual, the mirror is convex.

According to the convention

u=−18cm

m=1/3=−  

u

v

​  

 

⟹v=6cm

From mirror equation,

v

1

​  

+  

u

1

​  

=  

f

1

​  

 

⟹f=  

u+v

uv

​  

 

⟹f=9cm