Numerical Examples

Welcome class! Today we’ll discuss the lens formula. The formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Can anyone tell me what each symbol represents?

Is $f$ the focal length?

Exactly! And what about $v$ and $u$?

$v$ is the image distance, and $u$ is the object distance.

Great! Just remember, the distances are measured from the optical center of the lens. Let’s move to our first example!

In our first example, we have a convex lens forming an image at 20 cm when the object is placed at 30 cm. Can we identify the values of $u$ and $v$?

Yes, $u = -30 \, \text{cm}$ and $v = +20 \, \text{cm}$.

Perfect! Now, substituting these values into the lens formula, what do we get?

We calculate $\frac{1}{f} = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}$, hence $f = 12 \, \text{cm}$!

Well done! Remember, this shows how the convex lens converges light to a point. Let’s go to the next example.

Now, we’ll examine a concave lens. Here, we have an object placed 10 cm from the lens with a focal length of -15 cm. Can we set up our values for $u$ and $f$?

$u = -10 \, \text{cm}$ and $f = -15 \, \text{cm}$.

Excellent! Applying the lens formula $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$, what do you calculate?

We find $\frac{1}{v} = -\frac{1}{15} - \frac{1}{10} = -\frac{8}{30}$, leading to $v = -3.75 \, \text{cm}$!

Exactly! The negative value indicates a virtual image. Remember the characteristics of images formed by concave lenses!

Now let’s summarize the image characteristics for different types of lenses. Who remembers what happens with a convex lens?

It can produce real, inverted images that can be enlarged or diminished depending on the object's position!

Correct! And what about concave lenses?

Concave lenses always form virtual, erect, and diminished images.

Well done everyone! You’ve grasped important concepts of lenses. Remember, both types of lenses serve different applications.