Question

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

Answer 1

Given:

• u = -20 cm
• f = 15 cm.
• height of object = 5 cm.

To find:

• v
• height of image
• nature of image

Solution:

Using mirror formula,

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

∴  1/v = 1/15 - 1/-20

∴  1/v = 1/15 + 1/20

∴  1/v = 4+3/60

∴  1/v = 7/60

⇒  v = 8.5 cm.

Now we know that,

Magnification =  -v/u =  height of image/height of object

⇒  -60/7 × 20 = height of image/height of object

∴  height of image = 60/7×20 × 5

∴  height of image = 60/28

∴  height of image = 2.14 cm.

Hence the distance of image from the mirror = 8.5 cm

Nature of the image -  Virtual and Erect (as the mirror is convex)

Size of the image = 2.14 cm

Hope this helps!

Regards,

Soham Patil.

Answer 2

Answer

r=2f
=>f=r/2
radius of the curvature=30cm
then the focal lenght=15 [from avobe explanation]
the focal lenght of convex mirror is positive

2. Height of the object=5.0 or 5cm
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{20} = \\ \frac{1}{60} \\ = > v = 60$
$m = - \frac{v}{u} = \frac{60}{20} = 3$

size of the image will large 3×5=15cm
i have less time plz calculate distance own

hope other sulition will help u