Question

An object 5 cm in length is is placed at a distance of 20cm in front of convex mirror of radius of curvature 30cm find the position of the image its nature and size

Answer 1

Given :

• Object Height (ho) = 5 cm
• Image Distance (u) = - 20 cm
• Radius of curvature (R) = 30 cm
• Convex Mirror

To Find :

• Position, nature and size of the image

Solution :

As we know that,

➠ R = 2f

⇒30 = 2f

⇒f = 30/2

⇒f = 15

$\therefore$ Focal length of the mirror is 15 cm

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Now, use mirror formula :

➠1/f = 1/v + 1/u

⇒1/15 = 1/v + (-1/20)

⇒1/15 = 1/v - 1/20

⇒1/v = 1/15 + 1/20

⇒1/v = 4 + 3/60

⇒1/v = 7/60

⇒v = 60/7

⇒v = 8.57

$\therefore$ Image position is 8.57 cm from mirror

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Size and nature of Image can be find out using formula of Power of mirror,

➠ m = -v/u = h'/ho

⇒-8.57/-20 = ho/5

⇒ho = 0.437 * 5

⇒ ho = 2.185

$\therefore$ Height of object is 2.185 cm and nature is Virtual and erect

Answer 2

Given :

• Distance of object from mirror (u) = -20 cm
• Height of object (H₀) = 5 cm
• Radius of curvature (R) = 30 cm
• Focal length (F) = 15 cm

To Find :

We have to find the position,nature and size of the image

Explanation :

We know the mirror formula,

⇒ 1/F = 1/u + 1/v

→ 1/v = 1/F - 1/u

→ 1/v = 1/15 - (1/-20)

→ 1/v = 4 - (-3)/60

→ 1/v = 7/60

→ v = 60/7

→ v = 8.57

The image is formed at a distance of 8.57 cm behind the convex mirror. Thus, it is virtual and erect.

$\rule{200}{2}$

Now,

⇒ m = -v/u

→ m = -8.57/-20

→ m = 0.4

$\rule{150}{2}$

⇒ m = H₁/H₀

→ 0.4 = H₁/5

→ H₁ = 5 × 0.4

→ H₁ = 2 cm

Thus, the height of an image is 2 cm.  

Thus, we can say that that image is diminished.