Question 14). An object 5.0 cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size

Magnification =

\frac{I m a g e D i s \tan c e (H_{1})}{O b j e c t D i s \tan c e (H_{0})} = \frac{- 8.57}{- 20} = 0.428

Magnification =

\frac{H i g h t o f t h e i m a g e}{H i g h t o f t h e O b j e c t} = \frac{h}{h}

The positive value of image height indicates that the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller in size.

Question 14). An object 5.0 cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size- Hosting Precise

Posted on 4th Sep 2021 00:49:53 in

Question 14). An object 5.0 cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.

Ans:-

Object distance, u = −20 cm
Object height, ho = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length
R = 2f
f = 15 cm
According to the mirror formula,

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

= $\frac{1}{15} + \frac{1}{20} = \frac{7}{60}$

v = 8.57 cm

The positive value of v indicates that the image is formed behind the mirror.

Magnification = $\frac{I m a g e D i s \tan c e (H_{1})}{O b j e c t D i s \tan c e (H_{0})} = \frac{- 8.57}{- 20} = 0.428$

Magnification = $\frac{H i g h t o f t h e i m a g e}{H i g h t o f t h e O b j e c t} = \frac{h}{h}$

$h = m \times h$

=0.428*5

=2.14cm

The positive value of image height indicates that the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller in size.

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