Question

A diverging lens with \(f=-30.0 \mathrm{~cm}\) is placed \(15.0 \mathrm{~cm}\) behind a converging lens with \(f=20.0 \mathrm{~cm}\). Where will an object at infinity in front of the converging lens be focused?

Short answer

Answer: The object will be focused at a distance of \(5.0~cm\) behind the diverging lens.

Step-by-step solution

Find the image position of object due to the converging lens

Since the object is at infinity, the image will be at the focal point of the converging lens. So, the image position due to the converging lens (\(d_{i1}\)) is \(20.0~cm\) behind it.

Find the object position for the diverging lens

The object for the diverging lens will be the image formed by the converging lens. We know that the distance between the two lenses is \(15.0~cm\), so the object is \(20.0 - 15.0 = 5.0~cm\) away from the diverging lens. We can denote this distance as \(d_{o2}\).

Use Lensmaker's equation for the diverging lens

Lensmaker's equation is given as follows: \(\frac{1}{f} = \frac{1}{d_o} - \frac{1}{d_i}\) For the diverging lens, \(f = -30.0~cm\) and \(d_{o2} = 5.0~cm\). We will now use this equation to find the image position for the diverging lens (\(d_{i2}\)). \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\)

Solve for \(d_{i2}\)

We have the equation: \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\) To solve for \(d_{i2}\), we will first subtract \(\frac{1}{5.0}\) from both sides: \(\frac{1}{-30.0} - \frac{1}{5.0} = -\frac{1}{d_{i2}}\) Now we need to find a common denominator for the fractions on the left-hand side of the equation. The common denominator will be \(30.0\): \(\frac{-5-1}{30.0} = -\frac{1}{d_{i2}}\) \(\frac{-6}{30.0} = -\frac{1}{d_{i2}}\) Now we have fractions with the same numerator, so we can easily find \(d_{i2}\): \(d_{i2} = 30.0 \cdot \frac{1}{6} = 5.0~cm\)

Final Answer

The final image position is \(5.0~cm\) behind the diverging lens.