Question

Using diagram derive the lens maker formula.​

Answer 1

$\huge\rm\underline\purple{Question :-}$

Using diagram derive the lens maker formula.

$\huge{\underbrace{\textsf{\textbf{\color{lightblue}{Answer:-}}}}}$

From the above figure,

In ∆ABC,

$\frac{n1}{BO}$+$\frac{n2}{BI1}$= $\frac{n2-n1}{BC1}$

In ∆ADC,

$\frac{n2}{-DI1}$+$\frac{n1}{D1}$= $\frac{n1-n2}{-DC2}$

$\Large{\underline{\rm{\green{As \:the\: lens \:is\: very\: thin,}}}}$

$\Large{\underline{\rm{\green{B\:and\:D\: point\: can\: be\: taken \:as\: same.}}}}$

Therefore,

$\frac{n1}{BO}$+$\frac{n2}{BI1}$= $\frac{n2-n1}{BC1}$───────── ( i )

$\frac{n1}{BI}$-$\frac{n2}{BI1}$= $\frac{n2-n1}{BC2}$───────── ( ii )

Adding ( i ) and ( ii ),

$\frac{n1}{BI}$+$\frac{n1}{BO}$= n2-n1❴$\frac{1}{BC1}$+$\frac{1}{BC2}$❵

⇒$\frac{1}{BI}$+$\frac{1}{BO}$=$\frac{n2-n1}{nI}$ ❴$\frac{1}{BC1}$+$\frac{1}{BC2}$❵

Using sign convention,

• BI = v
• BC1 = R1
• BO = -u
• BC2 = -R2

⇒$\frac{1}{v}$+$\frac{1}{u}$=$\frac{n2-n1}{nI}$ ❴$\frac{1}{R1}$-$\frac{1}{R2}$❵

Now,

If u = ∝ and v = f,

⇒$\frac{1}{v}$-$\frac{1}{u}$=$\frac{1}{f}$

∴ $\frac{1}{f}$ = (n21 - 1)❴$\frac{1}{R1}$-$\frac{1}{R2}$❵

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Answer 2

Explanation:

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