- 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
(a) A small object is placed 150 mm away from a perging lens of focal length 100 mm.(i) Copy the figure below and draw rays to show how an image is formed by the lens.(ii) Calculate the distance of the image from the lens by using the lens formula.(b) The perging lens in part (a) is replaced by a converging lens also of focal length 100 mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the perging lens in part (a)."
(a)(i) The image formed by the lens is virtual, erect and diminished in size.
(a)(ii) Given: Diverging lens (Concave lens)
Object distance, $u$ = $-$150 mm = $-$15 cm (object distance always taken negative)
Focal length, $f$ = $-$100 mm = $-$10 cm (focal length of concave lens is always taken negative)
To find: Image distance, $v$.
Solution:
From the lens formula, we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values, we get-
$\frac {1}{v}-\frac {1}{(-15)}=\frac {1}{(-10)}$
$\frac {1}{v}+\frac {1}{15}=-\frac {1}{10}$
$\frac {1}{v}=-\frac {1}{10}-\frac {1}{15}$
$\frac {1}{v}=\frac {-3-2}{30}$
$\frac {1}{v}=-\frac {5}{30}$
$\frac {1}{v}=-\frac {1}{6}$
$v=-6cm=-60mm$
Thus, the distance of the image $v$ from the concave lens is 60 mm, and the positive $(-)$ sign implies that the image is formed in front of the lens (on the left side). And, we know that a virtual and erect image is formed in front of the lens.
(b) Given: Converging lens (Convex lens)
Object distance, $u$ = $-$150 mm = $-$15 cm (object distance always taken negative)
Focal length, $f$ = $+$100 mm = $+$10 cm (focal length of convex lens is always taken positive)
To find: Image distance, $v$.
Solution:
From the lens formula, we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values, we get-
$\frac {1}{v}-\frac {1}{(-15)}=\frac {1}{10}$
$\frac {1}{v}+\frac {1}{15}=\frac {1}{10}$
$\frac {1}{v}=\frac {1}{10}-\frac {1}{15}$
$\frac {1}{v}=\frac {3-2}{30}$
$\frac {1}{v}=\frac {1}{30}$
$v=+30cm=+300mm$
Thus, the distance of the image $v$ from the convex lens is 300 mm, and the positive $(+)$ sign implies that the image is formed behind the lens (on the right side). And, we know that a real and inverted image is formed behind the lens.
Comparison between the two properties of the image formed by the diverging lens and the image formed by the converging lens are as follow:
Diverging Lens (Concave Lens) | Diverging Lens (Concave Lens) |
1. Image formed is virtual and erect. | 1. Image formed is real and inverted. |
2. Size of the image formed is diminished (smaller than the object). | 2. Size of the image formed is magnified (larger than the object). |