A peanut is placed 40cmin front of a two-lens system: lens 1 (nearer the peanut) has focal length f₁ =20cm, lens 2 has f₂=-15cm and the lens separation is d=10cm. For the image produced by lens 2, what are (a) the image distance i₂(including sign), (b) the image orientation (inverted relative to the peanut or not inverted), and (c) the image type (real or virtual)? (d) What is the net lateral magnification?

1. Object distance, p₁=40cm.
2. Focal length of lens 1, f₁=20cm
3. Focal length of lens 2, f₂=-15cm
4. Lens separation, d=10cm

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

Formula:

The lens formula, $\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ $\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ (i)

The magnification formula of the lens, $m=-\frac{i}{p}$ (ii)

The overall lateral magnification of two lenses, m=m₁*m₂ (iii)

(a)

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

$$\begin{gathered} \frac{1}{40cm}+\frac{1}{i_1}=\frac{1}{20cm} \\ =\frac{1}{i_1}=\frac{1}{20cm}-\frac{1}{40cm} \\ =\frac{(40cm-20cm)}{\left(20cm\right)\left(40cm\right)} \\ =\frac{20cm}{800c{m}^2} \\ =\frac{1}{40cm} \\ {i}_1=40cm \end{gathered}$$

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

$$\begin{gathered} m_1=-\frac{40cm}{40cm} \\ =-1 \end{gathered}$$

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:

$$\begin{gathered} p_2=-40\mathrm{cm}+10\mathrm{cm} \\ =-30cm \end{gathered}$$

And the given focal length, f₂=-15cm

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

$$\begin{gathered} \frac{1}{-30cm}+\frac{1}{i_2}=\frac{1}{-15cm} \\ =\frac{1}{i_2}=\frac{1}{-15cm}-\frac{1}{-30cm} \\ =\frac{-30cm+15cm}{(-30cm)(-15cm)} \\ =\frac{-15cm}{450c{m}^2} \\ =\frac{-1}{30cm} \\ {i}_2=-30cm \end{gathered}$$

Thus, the image formed by the lens 2 is 30 to the left of lens 2.

Hence, the image distance is -30cm.

(b)

From equation (ii), 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:

$$\begin{gathered} m_2=-\frac{-30\mathrm{cm}}{-30\mathrm{cm}} \\ =-1 \end{gathered}$$

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

m=(-1)(-1)

=+1

The net magification is positive.

Hence, the image is not inverted.

(c)

The value of i₂ is negative.

Hence, the image is virtual.

(d)

The lateral magnification m of an object is given by, m=+1.

Hence, as calculated in part b, the overall magnification is +1.0.