- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall?

**Given:**

Height of the object, $h_{1}$ = 2 cm

Focal length of the mirror, $f$ = $-$10 cm

Height of the image, $h_{2}$ = 6 cm

**To find: **Distance of the object from the mirror, $u$.

**Solution:**

From the magnification formula, we know that-

$m=\frac{{h}_{2}}{{h}_{1}}=-\frac{v}{u}$

Substituting the given values in the magnification formula we get-

$\frac{6}{2}=-\frac{v}{u}$

$3=-\frac{v}{u}$

$v=-3u$

Now, from the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-10)}=\frac{1}{(-3u)}+\frac{1}{u}$

$\frac{1}{3u}-\frac{1}{u}=\frac{1}{10}$

$\frac{1-3}{3u}=\frac{1}{10}$

$\frac{-2}{3u}=\frac{1}{10}$

$3u=-20$

$u=\frac{-20}{3}$

$u=-6.67cm$

Thus, the distance of the object from the mirror $u$ is **-6.67 cm.**