Question

HOW TO EASILY DERIVE MIRROR AND LENS FORMULA ? NEED STEP BY STEP ANSWERS

Answer 1

Answer:

Derivation of formula for convex mirror:

Let AB be an object placed on the principal axis of a convex mirror of focal length f. u is the distance between the object and the mirror and v is the distance between the image and the mirror.

Convex mirror formula:

In ABC and A1B1C

<ABC = <A1B1C (right angles)

<ACB = <A1CB1

<CAB = <CA1B (common angle)

ABC is similar to A1B1C

AB/A1B1 = BC/B1C........(1)

similarly DEF issimilar to A1B1F

DE/A1B1 = EF/B1F....(2)

But DE = AB and when the aperture is very small EF = PF

Equation (2) becomes

AB/A1B1 = PF/B1F....(3)

Frm equations (1) and (3) get

PF/B1F = BC/B1C

PF/PF-PB1 = PB + PC/PC - PB1

f/f - v = -u + 2f/2f - v

[PF = f, PB1 = v, PB = u, PC = 2f]

2(2f - v)= (f-v)(2f-u)

-vf + uf + 2 fv -vu=0

fv+uf-vu=0....(4)

Dividing both sides of equation(4) by uvf we get

fv/uvf + uf/uvf - uv/uvf=0

1/u +1/v - 1/f=0

1/u + 1/v = 1/f

Answer 2

$\huge\sf\blue{Given}$

$\sf\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

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$\huge\sf\gray{To\;Prove}$

◈ The above statement

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$\huge\sf\purple{Steps}$

➳ $\sf \angle ACB = \angle A_1 CB_1$

Similarly,

➳ $\sf\angle ABC = \angle A_1B_1C_1 \: (Right \ Angles)$

As 2 Angles of the triangles are equal then obviously the third one should also be equal

➳ $\sf \angle BAC = \angle B_1A_1C$

And so,

➳ $\sf \dfrac{AB}{A_1B_1} = \dfrac{BC}{B_1C} \quad -eq(1)$

Similarly $\sf \triangle FED \ and \ FA_1B_1$ are also similar,so,

➳ $\sf \dfrac{ED}{A_1B_1} = \dfrac{EF}{FB_1}$

As ED = AB

➳ $\sf \dfrac{AB}{A_1B_1} = \dfrac{EF}{FB_1}\quad -eq(2)$

On Combining eq(1) and eq(2)

➳ $\sf \dfrac{BC}{B_1C} = \dfrac{EF}{FB_1}$

Now considered the point D is very close to P and so EF = PF

➳ $\sf\dfrac{BC}{B_1C} = \dfrac{PF}{FB_1}$

From the diagram we see that BC = PC-PB and $\sf B_1C = PB_1-PC$ and $\sf FB_1 = PB_1 - PF$

➳ $\sf \dfrac{PC-PB}{PB_1-PC} =\dfrac{PF}{PB_1-PF}$

$\quad$◕ PC = -R

$\quad$◕ PB = u

$\quad$◕ PB_1 = -v

$\quad$◕ PF = -f

Then the above Equation becomes,

➳ $\sf\dfrac{-R-(-u)}{-v-(-R)} =\dfrac{-f}{-v-(-f)}$

➳ $\sf\dfrac{u-R}{R-v} = \dfrac{-f}{f-v}$

➳ $\sf \dfrac{u-R}{R-v} = \dfrac{f}{v-f}$

On solving it,

➝ $\sf uv-uf-Rv+Rf = Rf-vf$

➝ $\sf uv-uf-Rv+vf=0$

Since R = 2f(Radius of curvature is twice that of the focal length)

➝ $\sf uv-uf-2fv+vf=0$

➝ $\sf uv-uf-fv = 9$

On solving ot further and finally dividing it with uv we have,

➝ $\sf\orange{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

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