- #1
- 84
- 0
The cartesian sign convention for a convex lens is follows:
1. All distances on the principal axis are measured from the optical center.
2. The distances measured in the direction of incident rays are positive and all the distances measured in the direction opposite to that of the incident rays are negative.
3. All distances measured above the principal axis are positive. Thus, height of an object and that of an erect image are positive and all distances measured below the principal axis are negative.
Then for a convex lens when the object is placed between 2f and f,the image is formed between pole and f on the other side.
The lens equation is 1/f=1/v-1/u.
In this case,
v,f>0 and u<0.
then 1/u=1/v-1/f=(f-v)/(vb)>0,.....BUT HOW CAN THIS BE TRUE as u<0.
1. All distances on the principal axis are measured from the optical center.
2. The distances measured in the direction of incident rays are positive and all the distances measured in the direction opposite to that of the incident rays are negative.
3. All distances measured above the principal axis are positive. Thus, height of an object and that of an erect image are positive and all distances measured below the principal axis are negative.
Then for a convex lens when the object is placed between 2f and f,the image is formed between pole and f on the other side.
The lens equation is 1/f=1/v-1/u.
In this case,
v,f>0 and u<0.
then 1/u=1/v-1/f=(f-v)/(vb)>0,.....BUT HOW CAN THIS BE TRUE as u<0.