An eraser of height1.0 cm is placed 10.0cmin front of a two-lens system. Lens 1 (nearer the eraser) has focallength, f₁=-15cm, lens 2 has f₂=12cm, and the lens separation is d=12cm. For the image produced by lens 2, what are (a) the image distance i₂(including sign), (b) the image height, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the eraser or not inverted)?

1. Object distance, p₁=10.0cm.
2. Focal length of lens 1, f₁=-15cm
3. Focal length of lens 2, f₂=12cm
4. Lens separation, d=12cm
5. Object height, h=1.0cm

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

Formula:

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

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

The overall lateral magnification of two lenses, m=m₁$\times$m₂ (iii)

The magnification of the lens considering the height of image and object,

$\left|m\right|=\frac{h^{\text{'}}}{h}$ (iv)

(a)

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:

$$\begin{gathered} \frac{1}{10cm}+\frac{1}{i_1}=\frac{1}{-15cm} \\ \frac{1}{i_1}=\frac{1}{-15cm}-\frac{1}{10cm} \\ =\frac{10cm+15cm}{(-15cm)\left(10cm\right)} \\ =\frac{25cm}{-150c{m}^2}=\frac{-1}{6cm} \\ {i}_1=-6cm \end{gathered}$$

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

$$\begin{gathered} m_1=-\frac{6.0\mathrm{cm}}{10\mathrm{cm}} \\ =0.60..\dots \dots \dots \dots \dots .\left(a\right) \end{gathered}$$

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:

p₂=12cm+6cm

=18cm

And the given focal length, f₂ =12cm

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

$$\begin{gathered} \frac{1}{18cm}+\frac{1}{i_2}=\frac{1}{12cm} \\ \frac{1}{i_2}=\frac{1}{12cm}-\frac{1}{18cm} \\ =\frac{18cm-12cm}{\left(12cm\right)\left(18cm\right)} \\ =\frac{6cm}{216c{m}^2}=\frac{1}{36cm} \end{gathered}$$

i₂=+36cm

Hence, the image distance is +36cm.

(b)

From equation (a), the magnification of lens 1 is found to be m₁. = -0.60.

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

$$\begin{gathered} m_2=-\frac{36\mathrm{cm}}{18\mathrm{cm}} \\ =-2 \end{gathered}$$

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

$$\begin{gathered} m=\left(0.60\right)\left(-2\right) \\ =-1.2 \end{gathered}$$

The image height can be calculated using the above data in equation (iv) as follows:

$\begin{array}{l}h\text{'}=\left|m\right|h\\ =(1.2)(1.0\mathrm{cm})\\ =1.2\mathrm{cm}\end{array}$

Hence, the value of the image height is 1.2cm.

(c)

The value of i₂ is positive.

Hence, the image is real.

(d)

The net magnification is negative.

Hence, the orientation of the final image is inverted with respect to the original object.