Question

$\huge\bold\pink{question}$


Two convex lenses of focal lengths 0.20 m and 0.30 m are kept
in contact. Find the focal length of the combination. Calculate
powers of two lenses and combinations?​

Answer 1

Answer :

Focal length of each lens is given as f1=0.20 m and f2=0.30 m.

Focal length of the combination F1=f11+f21

∴ F1=0.201+0.301

⟹ F=0.12 m

Power of lens P=f1

Power of first lens P1=0.201=5D

Power of second lens P2=0.301=3.33D

Power of combination of lens Pc=F1

⟹ Pc=0.121=8.33D

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Answer 2

$\large{\underline{\rm{\green{Solution:-}}}}$

$\implies{\boxed{\rm{P_{c} = \frac{1}{0.12} = 8.33D}}}$

$\large{\underline{\rm{\green{Explanation:-}}}}$

★ Given:-

• $d_{1} = 0.20m$
• $d_{2} = 0.30m$

★ To find:-

• Powers of two lenses & combination.

★ Now, WKT:-

Focal length of the combination, $\implies \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

$\implies \frac{1}{F} = \frac{1}{0.20} + \frac{1}{0.30}$

$\implies{\boxed{\rm{F = 0.12m}}}$

★ Using formula:-

Power of lens, $P = \dfrac {1}{f}$

Power of first lens, ${P}_{1} = \frac{1}{0.20}= 5D$

Power of second lens, ${P}_{2} = \frac{1}{0.30}= 3.33D$

Therefore, the powers of two lens are 5D and 3.33D.

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★ Now:-

Power of combination of lens, $P_{c} = \frac{1}{F}$

$\implies{\boxed{\rm{P_{c} = \frac{1}{0.12} = 8.33D}}}$

Hence, the power of combination of lens is 8.33D.

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