Physics, asked by ADITYAJNAMBIAR1, 5 hours ago

An object of height 4cm is placed on the principal axis at a distance of 20cm in front of a convex lens of focal length 30cm. Find the distance of image from the lens, its height and nature

Answers

Answered by nirman95
4

Given:

An object of height 4cm is placed on the principal axis at a distance of 20cm in front of a convex lens of focal length 30cm.

To find:

  • Image distance
  • Image Height
  • Image characteristics.

Calculation:

Applying Len's Formula:

 \dfrac{1}{f}  =  \dfrac{1}{v}  -  \dfrac{1}{u}

 \implies \dfrac{1}{30}  =  \dfrac{1}{v}  -  \dfrac{1}{( - 20)}

 \implies \dfrac{1}{30}  =  \dfrac{1}{v}  +  \dfrac{1}{ 20}

 \implies \dfrac{1}{v}  =  \dfrac{1}{30}  -  \dfrac{1}{20}

 \implies \dfrac{1}{v}  =  \dfrac{2 - 3}{60}

 \implies \dfrac{1}{v}  = -   \dfrac{1}{60}

 \implies \: v = -  60 \: cm

So, image distance is 60 cm.

  \implies \:  \dfrac{h_{i} }{h_{o} }  =  \dfrac{v}{u}

  \implies \:  \dfrac{h_{i} }{h_{o} }  =  \dfrac{ - 60}{ - 20}

  \implies \:  \dfrac{h_{i} }{h_{o} }  =  3

  \implies \: h_{i} =  3 \times 4 = 12 \: cm

So, image height is 12 cm.

Image characteristics:

  • Virtual
  • Erect
  • Magnified

Hope It Helps.

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