Question

80 through 87 80, 87 SSM WWW 83 Two-lens systems. In Fig. 34-45, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance. Lens 2 is mounted within the farther boxed region, at distance. Each problem in Table 34-9 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i₂for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification Mfor the system, including signs. Also, determine whether the final image is (c) real(R) or virtual(V), (d) inverted (I)from object O or non-inverted (NI), and (e) on the same side of lens 2 as object O or on the opposite side.

Short answer

1. The final image distance for the lens 2 is +10cm.
2. The overall lateral magnification of the system is -0.75.
3. The final image is real.
4. The final image is inverted.
5. The final image is formed on the opposite side of the lens 2 as O.

Step-by-step solution

The given data

• Object stands on the common central axis of two thin symmetric lenses.
• Distance between object and lens 1, p₁= +10cm.
• Distance between lens 1 and 2, d=10cm.
• Lens 1 is converging, focal length, f₁=15cm.

Lens 2 is converging, focal length, f₂=8cm.

Understanding the concept of properties of the lens

Using the relation between focal length, image distance, and object distance we can find the image distance. Then using the formula for overall magnification we can find its value. From the solution of parts a and b we can answer parts c, d, and e.

Formulae:

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 magnification of the two lens-systems,M=m₁m₂ (iii)

Here f is the focal length, i is the image distance, p is the object distance, mis the magnification.

a) Calculation of the final image distance for lens 2

The lens 1 is convergent, thus, the focal distance f₁ must be taken as positive. The object distance is given. So we can determine the distance of image1 from lens 1 using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1} \\ =\frac{1}{15}-\frac{1}{10} \\ =-\frac{1}{30} \\ {i}_1=-30cm \end{gathered}$$

Thus, image 1 is formed on the same side of lens 1 as O. It is virtual, non inverted and magnified.

This image 1 now acts as the object for the lens 2. The distance of this object from lens 2 will be sum of the image distance for lens 1 and the separation between the lenses as follows:

$\begin{array}{l}{p}_2=\left|i_1\right|+d\\ =30+10\\ =40\mathrm{cm}\end{array}$

Now the lens 2 is also convergent so its focal distance f₂ is also taken as positive.

We determine the image 2 distance i.e. final image distance from lens 2 is given using the above data in equation (i) as follows:

$$\begin{gathered} \frac{1}{i_2}=\frac{1}{f_2}-\frac{1}{p_2} \\ =\frac{1}{8.0}-\frac{1}{40} \\ =\frac{4}{40} \\ =\frac{1}{10} \\ {i}_2=+10cm \end{gathered}$$

Thus, the final image is formed at a distance +10cm from the lens 2.

b) Calculation of the overall lateral magnification

The overall lateral magnification is given using equation (ii) in equation (iii) with the given and calculated data from part (a) as follows:

$$\begin{gathered} M=\left(\frac{-i_1}{p_1}\right)\left(\frac{-i_2}{p_2}\right) \\ =\left(\frac{-(-30)}{10}\right)\left(\frac{-10}{40}\right) \\ =-0.75 \end{gathered}$$

Hence, the overall lateral magnification is -0.75.

c) Calculation of the property of the final image

Since the lens 2 is converging, the object for lens 2 is outside the focal point. The final image distance is positive.

Hence, the image formed by this lens system is real.

d) Calculation to know if the final image is inverted or not

Overall magnification for this lens system is negative which shows that the image and the object have the opposite orientation.

Hence, the image is inverted.

e) Calculation of the position of the final image

The final image distance is positive which shows that it is at positive side of the lens 2, which is on the opposite side of the object.

Hence, the final image is at the opposite side of the object for lens 2.