Question

12. An object is placed at a distance of 10 cm from a convex mirror of focal length
15 cm. Find the position and nature of the image.
​

Answer 1

Given

• Object distance (u) = - 10cm
• Focal length (f) = + 15cm

To find

• Position
• Nature of image

Solution

★Apply Mirror Formula★

$\implies\tt \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\tt \dfrac{1}{v}+\dfrac{(-1)}{10}=\dfrac{1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{3+2}{30} \\ \\ \\ \\ \implies\tt \dfrac{1}{v}=\cancel\dfrac{5}{30} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{1}{6} \\ \\ \\ \implies\tt v=+ 6cm$

Hence,

• Image distance = + 6cm
• Nature of image = Virtual and erect

$\rule{200}3$

Answer 2

Answer:

The image is obtained 6cm behind the mirror and is virtual and erect.

Explanation:

Given that,

u = -10cm

f = 15cm

According to mirror formula,

1/v + 1/u = 1/f

1/v - 1/10 = 1/15

1/v = 1/15 + 1/10

1/v = 5/30 = 1/6

Therefore v = 6cm.

Since the image distance is positive, we know that the image is virtual and erect.

I do so hope that this helps you.