Question

The sign conventions for a convex mirror are

a. v, f, u are positive b. v is positive, f & u are negative

c. u is positive, v and f are negative d. u is negative, v and f are positive​

Answer 1

Explanation:

The figure shows an object AB at a distance u from the pole of a concave mirror. The image A

1

B

1

is formed at a distance v from the mirror. The position of the image is obtained by drawing a ray diagram.

Consider the ΔA

1

CB

1

and ΔACB

∠A

1

CB

1

=∠ACB (vertically opposite angles)

∠AB

1

C=∠ABC (right angles)

∠B

1

A

1

C=∠BAC (third angle will also become equal)

∴ΔA

1

CB

1

and ΔACB are similar

∴

A

1

B

1

AB

=

B

1

C

BC

Similarly ΔFB

1

A

1

and ΔFED are similar

∴

A

1

B

1

ED

=

FB

1

EF

But ED=AB

A

1

B

1

AB

=

FB

1

EF

If D is very close to P then EF = PF

B

1

C

BC

=

FB

1

PF

BC=PC−PB

B

1

C=PB

1

−PC

FB

1

=PB

1

−PF

PB

1

−PC

PC−PB

=

PB

1

−PF

PF

But PC=R,PB=u,PB

1

=v,PF=f

Using sign convention,

PC=−R,PB=−u,PF=−fandPB

1

=−v

So we can equation (3) as :

−v−(−R)

−R−(−u)

=

−v−(−f)

−f

−v+R

−R+u

=

−v+f

−f

R−v

u−R

=

v−f

f

uv−uf−Rv+Rf=Rf−vf

uv−uf−Rv=Rf−Rf−vf

uv−uf−Rv=−vf

uv−uf−2fv=−vf (R = 2f)

uv−uf=2fv−fv

uv−uf=fv

Dividing throughout by uvf, we will get :

f

1

−

v

1

=

u

1

f

1

=

v

1

+

u

1

This is the required equation.