Question

plz help.............. ​

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{Height\:of\:object\:is\:5cm}\\\leadsto \bf{Object\:distance\:is\:10cm}\\\leadsto\bf{Focal\:length \:is\:10cm}\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto \bf{Location\:of\:image}\\\leadsto{\bf{Magnification\:produced}}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

We know, whenever in case if a concave lens if the object is at focus then the image is at infinity.

${\red{\underline{\sf{(\maps to Refer\:to\:ray\:diagram\:in\:attachment)}}}}$

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Let's use mirror's formula:

${\underline{\purple{\bf{\hookrightarrow Mirror\:Formula}}}}$

$\large\purple{\boxed{\boxed{\orange{\bf{\mapsto \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}}$

Using this,

$\sf{\implies \dfrac{1}{v}+\dfrac{-1}{10cm}=\dfrac{1}{-10cm}}$

$\sf{\implies \dfrac{1}{v}=\dfrac{1}{10cm}-\dfrac{1}{10cm}}$

$\sf{\implies \dfrac{1}{v}=\cancel{\dfrac{1}{10cm}}-\cancel{\dfrac{1}{10cm}}}$

$\sf{\implies \dfrac{1}{v}=0}$

$\sf{\implies v =\dfrac{1}{0}}$

${\underline{\boxed{\pink{\mapsto v=\infty}}}}$

${\bf{\red{\mapsto So\:object\:is\:formed\:at\:infinity}}}$

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Since the object is at infinity image will be highly enlarged.So,it's magnification will also be infinity.

$\bf{\green{\underline{\leadsto Magnification}}}$

$\large\purple{\boxed{\boxed{\orange{\bf{\mapsto m= \dfrac{f}{f-u}}}}}}$

$\sf{\implies m=\dfrac{10}{10-10}}$

$\sf{\implies m =\dfrac{10}{0}}$

${\underline{\boxed{\red{\mapsto m =\infty}}}}$

${\bf{\pink{\mapsto So\:magnification\:is\:infinity.}}}$

Answer 2

Answer:

$We know, whenever in case if a concave lens if the object is at focus then the image is at infinity.</p><p></p><p>{\red{\underline{\sf{(\maps to Refer\:to\:ray\:diagram\:in\:attachment)}}}}(\mapstoRefertoraydiagraminattachment)</p><p></p><p>______________________________________</p><p></p><p>Let's use mirror's formula:</p><p></p><p>{\underline{\purple{\bf{\hookrightarrow Mirror\:Formula}}}}↪MirrorFormula</p><p></p><p>\large\purple{\boxed{\boxed{\orange{\bf{\mapsto \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}}↦v1+u1=f1</p><p></p><p>Using this,</p><p></p><p>\sf{\implies \dfrac{1}{v}+\dfrac{-1}{10cm}=\dfrac{1}{-10cm}}⟹v1+10cm−1=−10cm1</p><p></p><p>\sf{\implies \dfrac{1}{v}=\dfrac{1}{10cm}-\dfrac{1}{10cm}}⟹v1=10cm1−10cm1</p><p></p><p>\sf{\implies \dfrac{1}{v}=\cancel{\dfrac{1}{10cm}}-\cancel{\dfrac{1}{10cm}}}⟹v1=10cm1−10cm1</p><p></p><p>\sf{\implies \dfrac{1}{v}=0}⟹v1=0</p><p></p><p>\sf{\implies v =\dfrac{1}{0}}⟹v=01</p><p></p><p>$

Step-by-step explanation:

sorry