Question

prove that for any spherical mirror 1/f=1/v +1/u,where the symbols have their usual meaning.​

Answer 1

Answer:

Explanation:

f= focal length....

v = image distance....

u= object distance.

Answer 2

Here in both proofs,

• the object OB of height x is placed and the image IM of height y is formed.  

• horizontal measurements along the direction of incident ray are taken as positive and opposite to the direction of the incident ray are taken as negative.

• vertical measurements above the principal axis are taken as positive and below the principal axis are taken as negative.

• The segment AP is neglected, such that AF = PF.

First consider the concave mirror. (Fig. 1)

The triangles OPB and IPM are similar since ∠OPB = ∠IPM = α (because the angle between the incident ray which passes through the pole and reflected ray thus formed is bisected by principal axis) and ∠POB = ∠PIM = 90°.

Thus,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{OB}{IM}=\dfrac{PO}{PI}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{x}{-y}=\dfrac{-u}{-v}} }$

$\small\text{ \displaystyle\sf{\longrightarrow-\dfrac{x}{y}=\dfrac{u}{v}\quad\quad\dots(1)} }$

The triangles AFC and IFM  are similar since ∠AFC = ∠IFM = β (alternating angles) and ∠FAC = ∠FIM = 90°.

Thus,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{AC}{IM}=\dfrac{AF}{FI}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{x}{-y}=\dfrac{-f}{-v+f}} }$

$\small\text{ \displaystyle\sf{\longrightarrow-\dfrac{x}{y}=\dfrac{f}{v-f}\quad\quad\dots(2)} }$

From (1) and (2) we get,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{u}{v}=\dfrac{f}{v-f}} }$

Now consider the convex mirror. (Fig. 2)

The triangles OPB and IPM are similar since ∠OPB = ∠IPM = α and ∠POB = ∠PIM = 90°.

Thus,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{OB}{IM}=\dfrac{OP}{PI}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{x}{y}=\dfrac{-u}{v}} }$

$\small\text{ \displaystyle\sf{\longrightarrow-\dfrac{x}{y}=\dfrac{u}{v}\quad\quad\dots(3)} }$

The triangles AFC and IFM  are similar since ∠AFC = ∠IFM = β and ∠FAC = ∠FIM = 90°.

Thus,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{AC}{IM}=\dfrac{AF}{IF}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{x}{y}=\dfrac{f}{f-v}} }$

$\small\text{ \displaystyle\sf{\longrightarrow-\dfrac{x}{y}=\dfrac{f}{v-f}\quad\quad\dots(4)} }$

From (3) and (4) we get,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{u}{v}=\dfrac{f}{v-f}} }$

This same equation appears in both the proofs and this equation leads to the following.

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{u}{v}=\dfrac{f}{v-f}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{v}{u}=\dfrac{v-f}{f}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{v}{u}=\dfrac{v}{f}-1} }$

Dividing each term by v,

$\small\text{ \displaystyle\sf{\longrightarrow\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}} }$

$\small\text{ \displaystyle\sf{\longrightarrow\underline{\underline{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}} }$

Hence the Proof!