91影视

• Image distance from the fly for lens 1,$\left|i_1+p_1\right|=20\text{cm}$
• Image distance from the fly for lens 2,$\left|i_2+p_2\right|=20\text{cm}$
• The height of the fly鈥檚 image,$H_1=2.0H$

Find the object distance from the lens 1. After that, you find the focal length of the lens 1. Again the object is situated in front of lens 2; then by using the concept of magnification, you find the image distance due to lens 2. After that, you can find the focal length from the lens 2.

Formulae:

The lens formula is,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ 鈥�(颈)

The magnification formula of the lens is,

$m=\frac{H\text{'}}{H}=-\frac{i}{p}$ 鈥�(颈颈)

If magnification is greater than one $\left(m_1>1\right)$, then the lens is converging, and for a converging lens, the focal length is positive.

From the given information and equation (ii), the relation of the image distance to the object distance can be given as follows:

localid="1663062400373" $\frac{2H}{H}=-\frac{i_1}{p_1}$

$i_1=-2p_1$ ...(iii)

Now, using the given data, you get the image and object distance for lens 1 as follows:

$$\begin{gathered} \left|p_1+i_1\right|=20\text{cm} \\ \left|p_1+(-2p_1)\right|=20\text{cm} \\ {p}_1=20\text{cm} \end{gathered}$$

Now, using this value in equation (iii), you get the image distance as:

$$\begin{gathered} i_1=-2脳20\text{cm} \\ =-40\text{cm} \end{gathered}$$

Substituting the above values into equation (I), you get the focal length of the lens 1 as follows:

$$\begin{gathered} f_1=\frac{p_1{i}_1}{p_1+i_1} \\ =\frac{\left(20\text{cm}\right)\left(-40\text{cm}\right)}{\left(20-40\right)\text{cm}} \\ =+40\text{cm} \end{gathered}$$

Here, the focal length is positive for a converging lens.

Hence, the focal length value is $+40\text{cm}$.

From result of part (a), the object distance $p_1$of the fly is $20\text{cm}$.

In this case, magnification is less than 1 and greater than zero and thus, the lens 2 is a diverging lens. So, the image produced by the lens 2 is virtual, and our result for the focal length is negative.

From the given information and equation (ii), the relation of the image distance to the object distance can be given as follows:

$\frac{H}{2H}=-\frac{i_2}{p_2}$

$i_2=-\frac{p_2}{2}$ 鈥�(颈惫)

Now, using the given data, you get the image and object distance for lens 2 as follows:

$$\begin{gathered} \left|p_2+-\frac{p_2}{2}\right|=20\text{cm} \\ \frac{p_2}{2}=20\text{cm} \\ {p}_2=40\text{cm} \end{gathered}$$

$$\begin{gathered} i_2=-\frac{40\text{cm}}{2} \\ =-20\text{cm} \end{gathered}$$

Substituting the above values into equation (i), you get the focal length of the lens 1 as follows:

$$\begin{gathered} f_2=\frac{i_2{p}_2}{p_2+i_2} \\ =\frac{\left(40\text{cm}\right)\left(-20\text{cm}\right)}{\left(40-20\right)\text{cm}} \\ =-\text{40 cm} \end{gathered}$$

Because focal length is negative, the lens is diverging.

Hence, the focal length of lens 2 is $-\text{40 cm}$.

From result of part (c), the object distance $p_2$of the fly is$40\text{cm}$