Question

An object of height 4 cm is placed in front of a convex lens of focal length 20 cm. If the object
ind the nature notion and stre of the image Calculate the magnitication.​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{QuestioN :}}}}}$

An object of size 4cm is place at a distance of 15cm in front of a concave mirror of focal length 20cm. What is the position, size and nature of the image? What is the magnification ?

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

GiveN :

• Object Height (ho) = 4 cm
• Object Distance (u) = - 15 cm
• Convex lens
• Focal length (f) = 20 cm

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To FinD :

• Size, position and nature of Image
• Magnification

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SolutioN :

We know that Lens formula is :

$\large{\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{u}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{20} \: - \: \dfrac{-1}{15}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{3 \: + \: 4}{60}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{7}{60}}} \\ \\ \implies {\sf{v \: = \: 8.5}}$

Image Distance is 8.5 cm

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As we know that :

Magnification = -v/u

⇒m = -8.5/-15

⇒m = 0.5

• Magnification is + 0.5
• Image size is diminished
• Nature is Virtual and erect