Question

A concave mirror has a focal length 20 cm. A rod of length 4 cm is placed along the principal axis of mirror in such a way that the end closer to lens lies on its centre of curvature. The length of image of rod is

Answer 1

if an object place on center of curvature then it image is also on center of curvature and height of image= height of the object image height is 4 cm and image distance is 2F=C=40cm

Answer 2

Answer:

Image  is 27.06 cm length.

Explanation:

Given,

A concave mirror has a focal lenghth 20 cm and a rod of length 4cm placed along principal axis of mirror in such a way that the end closer to lies on its centre of curvature.

So, we have  

Focal length of a concave mirror (f) = 20 cm

where  C (centre of curvature ) = 2f  (f is focal length)

⇒ 2 × 20 cm = 40 cm.    

Length of the rod is 4 cm .

Step 1:

Let AB be a rod of length = 4cm and given that end closer to lens lies on its centre of curvature  ,it means B line on its centre of curvature.

Image of B ,

$u_{1}$ = 40cm (distance  of B from the the mirror )

From mirror formula we have ,

         $\frac{1}{v_{1} } +\frac{1}{u_{1} } =\frac{1}{f}$

∴        $\frac{1}{v_{1} } +\frac{-1}{40} =\frac{-1}{20}$ ⇒    $\frac{1}{v_{1} } =\frac{-1}{20} +\frac{1}{40}$

         $\frac{1}{v_{1} } =\frac{-1}{40}$

       $v_{1}$ = -40cm (image of B)

Step2:

Image of A,

$u_{2}$ = 44cm  (distance of A from the mirror)

$\frac{1}{v_{2} } +\frac{1}{u_{2} } =\frac{1}{f}$

⇒$\frac{1}{v_{1} } +\frac{-1}{44} =\frac{-1}{20}$

⇒$\frac{1}{v_{1} } =\frac{-1}{20} +\frac{1}{44}$

⇒$\frac{1}{v_{1} } =\frac{22-5}{220}$ ⇒ $\frac{1}{v_{1} } =\frac{17}{220}$

  ⇒$v_{2} = 12.94$   (image of A)

Step 3:

Length of the rod AB

= (distance of the image A + distance of the image B)

= 12.94 +(-40) = -27.06cm

Final answer:

Hence the length of the image of rod is 27.06cm .