lens

[logistic_guy]

\(\displaystyle \bold{(a)}\) An object \(\displaystyle 37.5 \ \text{cm}\) in front of a certain lens is imaged \(\displaystyle 8.20 \ \text{cm}\) in front of that lens (on the same side as the object). What type of lens is this and what is its focal length? Is the image real or virtual? \(\displaystyle \bold{(b)}\) If the image were located, instead, \(\displaystyle 46.0 \ \text{cm}\) in front of the lens, what type of lens would it be and what focal length would it have?

 

[khansaheb]

\(\displaystyle \bold{(a)}\) An object \(\displaystyle 37.5 \ \text{cm}\) in front of a certain lens is imaged \(\displaystyle 8.20 \ \text{cm}\) in front of that lens (on the same side as the object). What type of lens is this and what is its focal length? Is the image real or virtual? \(\displaystyle \bold{(b)}\) If the image were located, instead, \(\displaystyle 46.0 \ \text{cm}\) in front of the lens, what type of lens would it be and what focal length would it have?

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[logistic_guy]

We can solve this problem by lens equation.

\(\displaystyle \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\)

You should not plug in values blindly in the equation. It is better to draw a diagram and be careful about where the image and the object are positioned. For example, because the image in this problem is in the same side as the object it is negative! And of course we can say directly it is virtual as it is in front of the lens.

Let us fill the lens equation with the given values.

\(\displaystyle \bold{(a)}\)

\(\displaystyle \frac{1}{37.5} + \frac{1}{-8.2} = \frac{1}{f}\)

This gives:

\(\displaystyle f = \textcolor{blue}{-10.5 \ \text{cm}}\)

It is \(\displaystyle \textcolor{blue}{\text{diverging}}\) as \(\displaystyle f < 0\).

It is \(\displaystyle \textcolor{blue}{\text{virtual}}\) as it is in front of the lens.


\(\displaystyle \bold{(b)}\)

\(\displaystyle \frac{1}{37.5} + \frac{1}{-46} = \frac{1}{f}\)

This gives:

\(\displaystyle f = \textcolor{blue}{203 \ \text{cm}}\)

It is \(\displaystyle \textcolor{blue}{\text{converging}}\) as \(\displaystyle f > 0\).

It is \(\displaystyle \textcolor{blue}{\text{virtual}}\) as it is in front of the lens.