An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Question

3 years ago

6

15 cm?

1 answer:

Lady_Fox [76] · Answer 1 · 2022-08-19T05:36:14+03:00

3 0

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

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${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

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

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$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

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$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

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$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

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$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

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$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

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<h3>☯ <u>Now, Finding the magnification </u></h3>

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$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

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$\dashrightarrow\:\: \sf{m = -2}$

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<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$