Question

An object is placed at distance 40 cm from convex mirror of focal length 20 cm. The image distance from the pole of a mirror is.​

Answer 1

Answer:

✡ 13.33cm ☑

Explanation:

Given ⤵

➡Object distance(u) = -40cm

➡Focal length (f) = 20

To find⤵

➡Image distance (v)

Solution⤵

⚫Using mirror formula

↪ $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$, we get

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

➡$\frac{1}{v}=\frac{1}{20}-\frac{1}{(-40)}$

➡$\frac{1}{v}=\frac{1×2}{20×2}-\frac{(-1)}{40}$

➡$\frac{1}{v}=\frac{2+1}{40}$

➡$\frac{1}{v}=\frac{3}{40}$

➡v= 40/3cm

➡v=13.33cm

✅Hence, the image distance is 13.33cm.

✔Extra points⤵

✡Mirror formula

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

Hope this is helpful to you!

Answer 2

Provided that:

• Distance of object = 40 cm
• Focal length = 20 cm

* Don't use the above information in your answer as it let your answer to be wrong as here we haven't use sign convention till now.

According to sign convention:

• Distance of object = -40 cm
• Focal length = +20 cm

To calculate:

• Distance of image

Solution:

• Distance of image = 13.33 cm

Using concept:

• Mirror formula

Using formula:

• Mirror formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}}}$

Where, v denotes image distance, f denotes focal length and, u denotes object distance.

Knowledge required:

• If the magnification produced by a spherical mirror is in negative then the mirror is always “Concave Mirror.”

• If the magnification produced by a spherical mirror is in positive then the mirror is always “Convex Mirror.”

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.

Required solution:

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{v} + \dfrac{1}{-40} = \dfrac{1}{20} \\ \\ :\implies \sf \dfrac{1}{v} - \dfrac{1}{40} = \dfrac{1}{20} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{40} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{2 \times 1 + 1 \times 1}{40} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{2 + 1}{40} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{3}{40} \\ \\ :\implies \sf 1 \times 40 = v \times 3 \\ \\ :\implies \sf 40 = 3v \\ \\ :\implies \sf \dfrac{40}{3} \: = v \\ \\ :\implies \sf 13.33 \: = v \\ \\ :\implies \sf v \: = 13.33 \: cm \\ \\ :\implies \sf Image \: distance \: = 13.33 \: cm$

_________________________

LCM of 40 & 20:

2 | 40 - 20

2 | 20 - 10

5 | 10 - 5

2 | 2 - 1

⠀| 1-1

→ 2 × 2 × 5 × 2

→ 4 × 5 × 2

→ 20 × 2

→ 40

_________________________

$\tiny{\begin{array}{c|c} \tt  \: \: 2 & \sf{40 - 20} \\ \\ \tt \: \:  2 & \sf {20 - 10} \\ \\ \tt 5 & \sf{10 - 5} \\ \\ \tt  \: \: 2 & \sf{2 - 1} \\ \\ \tt  \: \: 1 & \sf{1 - 1}\\ \\ \tt & \sf{1 - 1} \end{array}}$