Question

For a concave lens of focal length $f$, the relationship between object and image distances $u$ and $v$, respectively, from its pole can best be represented by ( $u=$ is the reference line)

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Step 1 : Given data

Focal length is $f$ [ $\because f<0$, As the lens is concave lens]

Object distance is $u$ $\left[\because u<0\right]$

Image distance is $v$ $\left[\because v<0\right]$

At reference line $u=v$

Step 2 : To find its pole best-represented graph

We know the lens formula

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ v=\frac{uf}{u+f} \end{gathered}$$

At reference line

$u=v$

$$\begin{gathered} v=\frac{(u+f-f)\times f}{u+f} \\ v=\left(\frac{(u+f-f)}{u+f}\right).f \\ v=\left(1-\frac{f}{u+f}\right).f \\ v=f-\frac{f^2}{u+f} \end{gathered}$$ [$f<0$]

If

$$\begin{gathered} u=\infty \\ v=f-\frac{f^2}{\infty } \\ v=f \end{gathered}$$

$v=f$for exact idea of curve slope at$u=0$ will be${45}^0$ Then, the slope is

$$\begin{gathered} \frac{dv}{du}=0+\frac{f^2}{{\left(u+f\right)}^2} \\ {\left[\frac{dv}{du}\right]}_{u=0}=1 \end{gathered}$$

Now, in this graph, the curve, $v=f$ meet at infinity and the tangent from origin is $u=v$ . The tangent should not cut the curve.

Hence, the correct answer is Option B.