Deduce the relation among object distance, image distance and refractive index for a convex spherical surfaces...
When rays are moving from denser to rarer medium forming virtual image
Answers
Answer:
good question I think this is the one
Explanation:
one medium (rarer) and on the right side the second medium (denser). P is the pole of the spherical surface and C is its centre of curvature. According to ray diagram of fig, I is the image of the object O. The normal drawn from point M on principal axis is MP’.
According to Snell's law,
1
n
2
=
sinr
sini
where
1
n
2
= the refractive index of medium (2) with respect to medium (1).
n=
sinr
sini
(on taking n in place of
1
n
2
temporarily ) If the angles i and r are very small, then
sini≈i and sin≈r
∴n=
r
i
or i=nr ...........(1)
∴ In triangle, the exterior angle is equal to sum of the opposite two interior angles.
∴ From ΔMOC,i=α+γ ..........(2)
And from ΔMIC,
r=β+γ ...........(3)
Substituting the value of i and r in equation (1), then we get,
(α+γ)=n(β+γ) ............(4)
If point M is not far from the principal axis, then
(i) Points P and P' are very close to each other and considered to be same point.
(ii) Angles α,β and γ will be small.
Therefore, α≈tanα=
OP
MP
=
−u
MP
β=tanβ=
IP
MP
=
−v
MP
and γ=tanγ=
CP
MP
=
R
MP
Now, substituting the values of α,β and γ in equation (4) then, we get
(
−u
MP
+
R
MP
)=n(
−v
MP
+
R
MP
)
or −
u
1
+
R
1
=−
v
n
+
R
n
or
v
n
−
u
1
=
R
n−1
Now again writing
1
n
2
in place of n, we get
v
1
n
2
−
u
1
=
R
1
n
2
−1
........(5)
This formula is known as 'Refraction Formula' of convex surface. If the absolute refractive indices of medium 1 and 2 be n
1
and n
2
respectively, then
1
n
2
=
n
1
n
2
∴ From equation (5)
n
1
v
n
2
−
u
1
=
R
n
1
n
1
−1
=
n
1
R
n
2
−n
1
or
v
n
2
−
u
n
1
=
R
n
2
−n
1
This equation is true for all types of spherical surfaces.
Answer:
Refer to the attachment
