A coin is placed 20cmin front of a two-lens system. Lens 1 (nearer the coin) has focal length f₁ =12cm, lens 2 has f₂=12.5cm, and the lens separation is d=30cm. For the image produced by lens 2, what are (a) the image distancei₂(including sign), (b) the overall lateral magnification, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the coin or not inverted)?

• Object distance, p₁ =20cm
• The focal length of lens 1, f₁=10cm
• The focal length of lens 2, f₂=12.5cm
• Lens separation, d=30cm

We use the lens equation to find the image distance. If image distance is positive, the image is real and if it is negative, the image is virtual. If the net magnification is negative, the final image will be inverted for the original object.

Formulae:

The lens formula is,

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$ ….. (i)

The magnification formula of the lens is,

$m=-\frac{i}{f}$ ….. (ii)

The overall lateral magnification of the two lenses,

$m=m_1\times m_2$

The mirror equation relates an object distance p₁ , mirrors focal length f₁ and the image distance i₁ due to the first mirror is given using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{20cm}+\frac{1}{i_1}=\frac{1}{10cm} \\ \frac{1}{i_1}=\frac{1}{10cm}-\frac{1}{20cm} \\ =\frac{20cm-10cm}{\left(20cm\right)\left(10cm\right)} \\ =\frac{10cm}{200c{m}^2} \\ =\frac{1}{20cm} \\ {i}_1=20cm \end{gathered}$$

The magnification of the image can be calculated using the above data in equation (ii) as follows:

$$\begin{gathered} m_1=-\frac{20cm}{20cm} \\ {m}_1=-1 \end{gathered}$$

This image will act as the role of the object for the second lens.

So, now for the second lens, the object distance is given as:

$$\begin{gathered} p_2=30\mathrm{cm}-20\mathrm{cm} \\ =10cm \end{gathered}$$

And the given focal length, f₂=10cm.

Image distance is given using the above data in equation (i) as follows:

$$\begin{gathered} \frac{1}{10cm}+\frac{1}{i_2}=\frac{1}{12.5cm} \\ \frac{1}{i_2}=\frac{1}{12.5cm}-\frac{1}{10cm} \\ =\frac{10cm-12.5cm}{\left(10cm\right)(12.5cm)} \\ =\frac{-2.5cm}{125c{m}^2} \\ =\frac{-1}{50cm} \\ {i}_2=-50cm \end{gathered}$$

Thus, the image formed by the lens 2 is 50cm to the left of lens 2, which means it coincides with the object.

Hence, the image distance is -50cm.

From equation (a), the magnification of lens 1 is found to be m₁ = -1.

Magnification for the image for lens 2 is given using the required data in equation (ii) as follows:

$$\begin{gathered} m_2=-\frac{-50cm}{10cm} \\ =5cm \end{gathered}$$

Now, the overall lateral magnification of the lenses can be calculated using the above data in equation (iii) as follows:

m = (-1)(5)

= - 5.0

Hence, the overall magnification is - 5.0 .

The value of i₂ is negative.

Hence, the image is virtual.

The lateral magnification m of an object calculated in part b is m = - 5.0.

The net magnification is negative.

Hence, the image is inverted