How can you increase optical power (or lens strength) for diverging lens?

Answer 1

First, let's look at the equation for power: P = 1/f. This is true for both diverging and converging lenses.

However, diverging lens have "negative" power. This is because the focal length of a diverging lens is to the left of the lens and anything left of the lens is considered negative (just by convention).

This is where the question will try to trick you. If P = 1/(-f) for diverging lenses, then one would think a greater f would lead to greater power. For example, f=1 vs f=100 leads to a power of -1 and -0.01 respectively. It seems like the power of the f=100 is greater because -0.01 > -1, right?

However, power is simply just "negative" for diverging lens, once again by convention. So a decrease in f actually leads to greater power! The more negative the power is for a diverging lens, the greater it is.

The MCAT loves to trip people up with these kinds of questions -- ultimately decreasing the focal length increases power for both converging and diverging lens (they just have opposite signs).

Hope this helped :).