Question

When a converging lens is placed between an object and a screen , an image of height 4 cm is produced on the screen. Without changing the distance between the object and the screen , another image of height 16 cm is produced on the screen at another position of the lens. Find the height of the object.

Answer 1

Assume the distance between the object and screen is x, then object distance (between object and lens) plus image distance (between screen and lens) equals x:

$$x=i_1+o_1=i_2+o_2\space\space\space(0)$$

for both cases.

Also, we know that magnification is the image distance divided by object distanceб or image height divided by object height:

$$m_1=\frac{i_1}{o_1},\space m_1=\frac{h_{i1}}{h_o}=\frac{4}{h_o}\rightarrow\frac{i_1}{o_1}=\frac{4}{h}.\\\space\\ m_2=\frac{i_2}{o_2},\space m_2=\frac{h_{i2}}{h_o}=\frac{16}{h_o}\rightarrow\frac{i_2}{o_2}=\frac{16}{h}.$$

Divide one by another:

$$\frac{i_1o_2}{i_2o_1}=\frac{1}{4}\rightleftarrows \frac{i_1}{o_1}=\frac{i_2}{4o_2}.\space(1)$$

Multiply one by another:

$$\frac{i_1i_2}{o_1o_2}=\frac{64}{h^2}.\space(2)$$

Moreover, according to thin lens equation, we have

$$\frac{1}{f}=\frac{1}{i_1}+\frac{1}{o_1}=\frac{1}{i_2}+\frac{1}{o_2},\\\space\\ \text{or}\\\space\\ \frac{i_1+o_1}{i_1o_1}=\frac{i_2+o_2}{i_2o_2}\rightleftarrows \frac{i_1}{o_2}=\frac{i_2}{o_1}.\space(3)$$

Plug equation (1) in (2):

$$\frac{i_2^2}{4o_2^2}=\frac{64}{h^2}.\space(4)$$

Plug equation (3) in (2):

$$\frac{i_2^2}{o_1^2}=\frac{64}{h^2}.\space(5)$$

Since right parts of (4) and (5) are equal, the left pars are equal as well:

$$\frac{i_2^2}{4o_2^2}=\frac{i_2^2}{o_1^2}\rightarrow o_1=2o_2.$$

Now plug this o-equation in (3):

$$\frac{i_1}{o_2}=\frac{i_2}{2o_2}\rightarrow i_1=\frac{i_2}{2}.\space(6)$$

Now look at the equation (0) in the very beginning and plug (6) into it:

$$\frac{i_2}{2}+2o_2=i_2+o_2\rightarrow i_2=2o_2.\space(7)$$

Finally, it's time to plug (7) in (4):

$$\frac{(2o_2)^2}{4o_2^2}=\frac{64}{h^2},\\\space\\ 1=\frac{64}{h^2}\rightarrow h=8\text{ cm}.$$