Physics, asked by shivam30011970genius, 9 months ago

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10cm. the distance of the object from the lens is 15cm. Find the nature and position and size of the image . Also find its magnification​

Answers

Answered by Anonymous
30

\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}

\small{\underline{\blue{\sf{Given :}}}}

  • Object Height (h) = 2 cm
  • Focal Length (f) = 10 cm
  • Object Distance (u) = 15 cm
  • Convex Lens

\rule{200}{1}

\small{\underline{\green{\sf{Solution :}}}}

Lens Formula :

\large \star {\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{u}}}}  \\ \\  \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: + \: \dfrac{1}{u}}} \\ \\  \footnotesize {\sf{\pink{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Putting \: Values}}} \\ \\  \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{10} \: + \: \dfrac{1}{-15}}}  \\ \\   \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{3 \: - \: 2}{30}}}  \\ \\  \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{30}}}  \\ \\  \implies {\sf{v \: = \: 30 \: cm}} \\ \\ \large{\boxed{\sf{Image \: Position  \: = \: 30 \: cm}}}

\rule{200}{1}

Formula for Magnification is :

\large \star {\boxed{\sf{m \: = \: \dfrac{v}{u} \: = \: \dfrac{h'}{h}}}}

\implies {\sf{\dfrac{30}{-15} \: = \: \dfrac{h'}{2}}}

\implies {\sf{-2 \: = \: \dfrac{h'}{2}}}

\implies {\sf{h' \: = \: -2 \: \times \: 2}}

\implies {\sf{h' \: = \: -4 \: cm}}

  • Height is -4 cm
  • Nature is Real And Inverted
  • Distance is 30 cm
Answered by Decksboy
1

Answer:

Here

Explanation:

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