Question

In a lens displacement method, the separation between source and screen is $D = 18$ cm. Ratio of magnification for two positions of lens is 4, then focal length of lens is ______ cm

Answer 1

4

Answer 2

The lens displacement method is based on the principle that for a fixed distance $D$ between an object and a screen, there are generally two positions of a convex lens for which a real image is formed on the screen.

Let $D$ be the separation between the source (object) and the screen.
Let $u_1$ and $v_1$ be the object and image distances for the first position of the lens, respectively.
Then, we have:
$u_1 + v_1 = D$ (1)
The magnification for the first position is $m_1 = \frac{v_1}{u_1}$.

For the second position of the lens, due to the principle of reversibility of light, the object distance becomes $v_1$ and the image distance becomes $u_1$.
The magnification for the second position is $m_2 = \frac{u_1}{v_1}$.

From the expressions for $m_1$ and $m_2$, we can see that:
$m_1 m_2 = \left(\frac{v_1}{u_1}\right) \left(\frac{u_1}{v_1}\right) = 1$.

Given that the ratio of magnification for two positions of the lens is 4. Let's assume $\frac{m_1}{m_2} = 4$.
Since $m_1 m_2 = 1$, we can substitute $m_1 = \frac{1}{m_2}$ into the ratio equation:
$\frac{1/m_2}{m_2} = 4$
$\frac{1}{m_2^2} = 4$
$m_2^2 = \frac{1}{4}$
Taking the positive root (as magnification usually refers to the magnitude of the ratio of sizes or distances):
$m_2 = \frac{1}{2}$.

Now, we can find $m_1$:
$m_1 = \frac{1}{m_2} = \frac{1}{1/2} = 2$.
So, the two magnifications are 2 and 1/2.

Let's use the first magnification $m_1 = 2$.
$m_1 = \frac{v_1}{u_1} = 2 \implies v_1 = 2u_1$.

We are given $D = 18$ cm.
Substitute $v_1 = 2u_1$ into equation (1):
$u_1 + 2u_1 = 18$
$3u_1 = 18$
$u_1 = \frac{18}{3} = 6$ cm.

Now find $v_1$:
$v_1 = 2u_1 = 2 \times 6 = 12$ cm.

Finally, use the lens formula to find the focal length $f$:
$\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}$ (for a real object and real image, using magnitudes)
$\frac{1}{f} = \frac{1}{12} + \frac{1}{6}$
To add these fractions, find a common denominator, which is 12:
$\frac{1}{f} = \frac{1}{12} + \frac{2}{12}$
$\frac{1}{f} = \frac{1+2}{12}$
$\frac{1}{f} = \frac{3}{12}$
$\frac{1}{f} = \frac{1}{4}$
$f = 4$ cm.

Alternatively, we can find the displacement $x$ between the two lens positions.
$x = |v_1 - u_1| = |12 - 6| = 6$ cm.
The focal length can also be calculated using the formula:
$f = \frac{D^2 - x^2}{4D}$
$f = \frac{18^2 - 6^2}{4 \times 18}$
$f = \frac{324 - 36}{72}$
$f = \frac{288}{72}$
$f = 4$ cm.

Both methods yield the same result.