Question

the power of a combination of two lenses in contact is + 1.0 D. if the focal length of one of the lens is a combination is + 20.0 cm, the focal length of the other lens would be:
a) - 120.0 cm
b) + 80.0 cm
c) - 25.0 cm
d) - 20.0 cm​

Answer 1

Answer :-

Focal length of the other lens is -25 cm . [Option.c]

Explanation :-

We have :-

→ Power of the combination (P) = + 1 D

→ Focal length of one lens (f₁) = + 20 cm

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Firstly, let's calculate focal length of the combination of lenses .

f = 1/P

⇒ f = 1/1

⇒ f = 1 m

⇒ f = 100 cm

Now, we shall put values in the formula of focal length of a cmbination of lenses (equivalent focal length) .

1/f₁ + 1/f₂ = 1/f

⇒ 1/20 + 1/f₂ = 1/100

⇒ (f₂ + 20)/20f₂ = 1/100

⇒ 100(f₂ + 20) = 20f₂

⇒ 100f₂ + 2000 = 20f₂

⇒ -80f₂ = 2000

⇒ f₂ = 2000/-80

⇒ f₂ = -25 cm

Answer 2

Given :

• Focal length of combination of lenses = + 1.0 D
• Focal length of 1 lens = + 20.0 cm

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To Find :

• Focal length of other lens = ?

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Solution :

~ Formula Used :

• Focal Length(Combination of two lenses) :

$\large{\color{cyan}{\bigstar}} \; \; {\underline{\boxed{\red{\sf{ f = \dfrac{1}{p} }}}}}$

Where :

• f = focal length
• p = power of lenses

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~ Calculating the focal length of lenses :

$\begin{gathered} \; :\longmapsto \; \; \sf { f = \dfrac{1}{p} } \\ \end{gathered}$

$\begin{gathered} \; :\longmapsto \; \; \sf { f = \dfrac{1}{1} } \\ \end{gathered}$

$\begin{gathered} \; :\longmapsto \; \; {\qquad{\orange{\sf { f = 1 \; m \; or \; 100 \; cm }}}} \\ \end{gathered}$

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~ Calculating the focal length of 2nd lens :

$\begin{gathered} \; \dashrightarrow \; \; \sf { \dfrac{1}{f}_1 + \dfrac{1}{f}_2 = \dfrac{1}{f} } \\ \end{gathered}$

Where :

• ${\sf{\dfrac{1}{f}_1 }}$ = Focal length of 1st lens

• ${\sf{\dfrac{1}{f}_2 }}$ = Focal length of 2nd lens

• ${\sf{\dfrac{1}{f} }}$ = Focal length of Combination of lenses

$\begin{gathered} \; \dashrightarrow \; \; \sf { \dfrac{1}{f}_1 + \dfrac{1}{f}_2 = \dfrac{1}{f} } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { \dfrac{1}{20} + \dfrac{1}{f}_2 = \dfrac{1}{100} } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { \dfrac{f_2 + 20}{20 f_2} = \dfrac{1}{100} } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { 100(f_2 + 20) = 1 \times 20 f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { 100 f_2 + 2000 = 20 f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { 2000 = 20 f_2 - 100 f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { 2000 = - 80 f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { \dfrac{2000}{-80} = f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; \sf { \cancel\dfrac{2000}{-80} = f_2 } \\ \end{gathered}$

$\begin{gathered} \; \dashrightarrow \; \; {\qquad{\green{\sf { f_2 = - 25.0 \; cm }}}} \\ \end{gathered}$

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~ Therefore :

❛❛ Focal length of the other lens is -25.0 cm .So, C Option is correct . ❜❜

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