Q91P - University Physics with Modern Physics

Question

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

Step-by-Step Solution

Answer

The focal length of the diverging lens is -26.7cm.

Step 1:Relationship between object and image distance.

Object’s distance is s and image distance is s’.

The relationship is  $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}}$

Step 2:Calculation of the focal length.

Given the distance between the converging lens and the screen is 30cm and a diverging lens is placed 15cm to the right of converging lens

Therefore ,

The distance between the virtual object and diverging lens is (15-30)cm=-15cm.

The screen has to be moved 19.2cm, therefore s’=15cm+19.2cm=34.2cm

Now applying the formula,

$$\begin{gathered} \frac{1}{\mathit{f}}=\frac{1}{s}+\frac{1}{s\text{'}} \\ \mathit{f}=\frac{ss}{s+s\text{'}} \\ =\frac{\left(-15.0\right)\times 34.2}{-15.0+34.2} \\ =\frac{-513}{19.2} \\ =-26.7\mathbf{cm} \end{gathered}$$ 

Here the negative sign implies that the lens is diverging.