Physics, asked by mahimamahima9695, 8 months ago

An object 3cm high is place at a distance of 15cm from a concave mirror the radius of curvature is 20cm find the nature position and size of the image

Answers

Answered by maddulamounika111
4

Answer:

Explanation:

R=2f

f=10cm

from mirror equation,

1/f=1/v+1/u

1/10=1/v+1/3

Solve it to get image distance.

When the object placed between f and c then image formed beyond c

And the image is real and inverrted,and large than object..........

Answered by Anonymous
10

S  O L U T I O N :

\bf{\large{\underline{\bf{Given\::}}}}}

  • An object height (h1) = 3 cm.
  • The radius of curvature (R) = 20 cm.
  • Distance of object from mirror (u) = -15 cm.

\bf{\large{\underline{\bf{To\:find\::}}}}}

The nature position & size of the image.

\bf{\large{\underline{\bf{Explanation\::}}}}}

We know that formula of the focal length :

\longrightarrow\sf{Focal\:length\:(f)=\dfrac{Radius\:of\:curvature}{2} }\\\\\\\longrightarrow\sf{Focal\:length\:(f)=\cancel{\dfrac{-20}{2}} }\\\\\\\longrightarrow\sf{Focal\:length\:(f)=-10\:cm}

Now;

By using formula of mirror :

\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{v} +\dfrac{1}{u} }\\\\\\\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} +\bigg(\dfrac{1}{-15} \bigg)}\\\\\\\longrightarrow\sf{\dfrac{1}{-10}=\dfrac{1}{v}-\dfrac{1}{15} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{15} }\\\\\\\longrightarrow\sf{\dfrac{1}{v}=\dfrac{-3+2}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v}=\dfrac{-1}{30}}\\\\\\\longrightarrow\bf{v=-30\:cm}

∴ Image is formed at a distance of 30 cm in front of the mirror.

We know that formula of the magnification:

\mapsto\bf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image\:(v)}{Distance\:of\:object\:(u)}}

\longrightarrow\sf{m=\dfrac{h_2}{h_1} =\dfrac{-v}{u} }\\\\\\\longrightarrow\sf{\dfrac{h_2}{3} =\dfrac{-30}{-15} }\\\\\\\longrightarrow\sf{-15h_2=-30\times 3}\\\\\\\longrightarrow\sf{-15h_2=-90}\\\\\\\longrightarrow\sf{h_2=\cancel{\dfrac{-90}{-15} }}\\\\\\\longrightarrow\bf{h_2=6\:cm}

∴ Height of image is 6 cm, the image is real and Inverted.

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