draw a ray diagram to show the nature of image formed by a concave mirror of focal length 15 cm and distance of object is 20 cm
Answers
Answer:
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.
(c) (i) Reflecting telescopes do not suffer from chromatic aberration.
(ii) Reflecting telescope are cheaper to make than refracting telescopes of the same size
solution