Question 1

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 15 cm?

Question 2

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

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

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

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

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

$\\$

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

$\\$

$\dashrightarrow\:\: {\boxed{\sf{(1)/(u) + (1)/(v) = (1)/(f)}}}$

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$\dashrightarrow\:\: \sf{(1)/(v)-(1)/(-15)=(1)/(10)}$

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$\dashrightarrow\:\: \sf{(1)/(v)+(1)/(15)=(1)/(10)}$

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$\dashrightarrow\:\: \sf{(1)/(v) = (1)/(10) - (1)/(15)}$

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$\dashrightarrow\:\: \sf{(1)/(v) = (1)/(30)}$

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

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$\\$

$\dashrightarrow\:\: \sf{ m = (-30)/(-15)}$

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

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

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

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