Physics, asked by sk715rocks, 11 months ago

An object of 5 cm is placed 20
in front of a convex mirror of
radius30 cm Find the location, magnification and height of the image​

Answers

Answered by Anonymous
6

\large\underline{\bigstar \: \: {\sf Given-}}

In Convex Lens -

  • Object is at (u) = - 20cm
  • Radius of Curvature (R) = 30cm
  • Height of object {\sf h_o} = 5cm

Object Distance is negative Because it is at left side of mirror.

\large\underline{\bigstar \: \: {\sf To \: Find -}}

  • Location of Image
  • Magnification or image
  • Height of image

\large\underline{\bigstar \: \: {\sf Formula \: Used -}}

\implies\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}

\implies\underline{\boxed{\sf Magnification  \: (m)=\dfrac{-v}{u}}}

\implies\underline{\boxed{\sf m=\dfrac{h_i}{h_o}}}

\large\underline{\bigstar \: \: {\sf Solution-}}

\implies{\sf \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}

We know that -

\implies{\bf R=2f}

\implies{\sf 30=2f}

\implies{\sf f=\dfrac{30}{2}}

\implies{\sf f=15cm }

\bullet\underline{\bf Location \; of \: image -}

\implies{\sf \dfrac{1}{15}=\dfrac{1}{v}+\left(-\dfrac{1}{20}\right)}

\implies{\sf \dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{20} }

\implies{\sf \dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{20}}

\implies{\sf \dfrac{1}{v}=\dfrac{5+20}{20×5}}

\implies{\sf \dfrac{1}{v}=\dfrac{25}{100} }

\implies{\sf v=\dfrac{100}{25} }

\implies{\bf v=4\:cm}

\bullet Any position of object in a Convex mirror image is formed always between Pole (P) and Focus (F)

\bullet Behind the Mirror

\bullet Small, erect & virtual

\bullet\underline{\bf Magnification \: of \; image -}

\implies\underline{\boxed{\sf m=\dfrac{-v}{u}}}

\implies{\sf m=\dfrac{-4}{5} }

\implies{\bf m=-0.8}

Magnification of image is -0.8

\bullet\underline{\sf Height \; of \; image -}

\implies\underline{\boxed{\sf m=\dfrac{h_i}{h_o}}  }

{\sf h_i= Height \; of \: image }

{\sf h_o=Height \; of \: object}

\implies{\sf 0.8=\dfrac{h_i}{5} }

\implies{\sf h_i=5×0.8 }

\implies{\bf h_i=4\:cm }

Height of image is 4cm

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