Consider a concave mirror and a convex lens (refractive
index = 1.5) of focal length 10 cm each, separated by a
distance of 50 cm in air (refractive index = 1) as shown in
the figure. An object is placed at a distance of 15 cm from
the mirror. Its erect image formed by this combination has
magnification M₁. When the set- up is kept in a medium of
refractive index 7/6, the magnification becomes M₂. The
magnitude M₂/M₁ is
Answers
Explanation:
ANSWER
The first image from reflection by spherical concave mirror can be found from mirror formula:
v
1
+
u
1
=
f
1
...(i)
For concave surface: u=-15 cm, f=-10 cm
Substituting in equation (i)
v
1
+
−15
1
=
−10
1
=
v
1
=
150
10−15
⇒v=−30cm
the magnification is given as m=−
u
v
=−
−15
−30
=−2
The first image will act as object for lens and it's position can be found using lens formula
v
1
−
u
1
=
f
1
...(ii)
For concave surface: u=-20 cm, f=10 cm
Substituting in equation (ii)
v
1
−
−20
1
=
10
1
⇒
v
1
10
1
−
20
1
=
20
2−1
⇒v=20cm
the magnification is given as m=
u
v
=
−20
20
=−1
total magnificationM
1
=−2×−1=2
Now the whole set is immersed in liquid of n=
6
7
the reflection from concave mirror will remain same as their is no change in object position and focal length.
The change in focal length can be found using lens maker formula
f
1
=(
μ
1
μ
2
−1)(
R
1
1
−
R
2
1
)
f
air
1
=(1.5−1)(
R
1
1
−
R
2
1
)...(iii)
f
liquid
1
=(
6
7
1.5
−1)(
R
1
1
−
R
2
1
)...(iv)
dividing (iii) by (iv)
f
air
f
liquid
=
2
3
7
6
−1
2
1
f
air
f
liquid
=
2
3
7
6
−1
2
1
=
4
7
f
liquid
=f
air
×
4
7
=
4
70
The first image will act as object for lens and it's position can be found using lens formula
v
1
−
u
1
=
f
1
...(i)
For concave surface: u=−20cm,f=
4
70
cm
Substituting in equation (i)
v
1
−
−20
1
=
70
4
⇒
v
1
70
4
−
20
1
=
1400
80−70
⇒v=140cm
the magnification is given as m=
u
v
=
−20
20
=−7
total magnificationM
2
=−2×−7=14
M
1
M
2
=7
Answer:
Wait for 5min its answer is very easy.