Chapter 23, Problem 77GP

To determine

we have to find the focal length of the lens.

Answer to Problem 77GP

Solution: For determining the focal length of given lens we will use the lens formula.

Explanation of Solution

Given that an object is placed 60 cm from a certain converging lens (convex lens), it forms a real image. And when the object is moved closer to lens at 40 cm, the image moves 10 cm farther from the lens.

Given:

Given that in first case object distance from lens (say)$u_1=-\text{ 60 cm}$ (according to the sign convention for convex lens, when object is behind lens)

In second condition object distance from lens (say)$u_2\text{= - 40 cm}$ (according to sign convention of convex lens, when object is behind lens)

Let’s assume that focal length of lens is (say)$f$

And in first case image is formed at distance = $v_1$

So, in second case image distance from object would be = $\left(v_1+10\right)\text{cm}$

Formula used:

Focal length of convex lens if represented by

$1/f=1/v-1/u$

Calculation:

For the first case

$\begin{array}{l}1/f=1/v_1–1/u_1\\ 1/f=1/v_1-1/\left(-60\right)\\ 1/f=1/v_1+1/60\left(\text{equation 1}\right)\end{array}$

For the second case

$\begin{array}{l}1/f=1/\left(v_1+10\right)–1/\left(-40\right)\\ 1/f=1/\left(v_1+10\right)+1/40\left(\text{equation}\text{2}\right)\end{array}$

From equation 1 and 2, equating both we get

$\begin{array}{l}1/\left(v_1+10\right)+1/40=1/v_1+1/60\hfill \\ 1/\left(v_1+10\right)–1/v_1=1/60–1/40\hfill \\ \left(v_1–\left(10+v_1\right)\right)/v_1\left(v_1+10\right)=\left(2-3\right)/120\hfill \\ -10/v_1\left(v_1+10\right)=-1/120\hfill \end{array}$

$\begin{array}{l}{v}_1\left(v_1+10\right)=1200\hfill \\ v_1{}^2+10v_1–1200=0\hfill \\ v_1{}^2+40v_1-30v_1+1200=0 \left(\text{after factorizing}\right)\hfill \\ \left(v_1+40\right)\left(v_1-30\right)=0\hfill \\ v_1=-40or30\hfill \end{array}$

Now since we have got value of v₁ then by putting this value in equation 1 or 2 we can find the value of focal length f.

So by putting both the values of v₁ in equation 1, we find

$\begin{array}{l}\left(1\right) \text{Taking}{v}_1=30\\ 1/f=1/30+1/60=\left(2+1\right)/60=3/60f=60/3=20\text{cm}\\ \left(2\right)\text{ Taking}{v}_1=-40\text{ cm}\\ \begin{array}{l}\begin{array}{l}\\ 1/f=-1/40+1/60=\left(-3+2\right)/120=-1/120\end{array}\hfill \\ f=-120\text{cm}\hfill \end{array}\end{array}$

Since focal length of convex lens can not be negative so focal length of the lens is 20 cm.