Question

A converging lens has a focal length of 15 cm. The difference between the two object positions so that the size of the image is twice as large as the object is 15 cm.

Answer 1

15

Answer 2

To find the object positions for which the image size is twice the object size, we use the lens formula and magnification formula.

Given: Focal length of converging lens, $f = +15 \text{ cm}$ (positive for a converging lens). Magnitude of magnification, $|m| = 2$.

The magnification ($m$) for a lens is given by: $m = \frac{v}{u}$ And the lens formula is: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

From the magnification formula, we can write $v = mu$. Substitute this into the lens formula: $\frac{1}{mu} - \frac{1}{u} = \frac{1}{f}$ Multiply by $mu$ to clear denominators: $1 - m = \frac{mu}{f}$ Rearranging to solve for $u$: $u = \frac{f(1 - m)}{m}$

There are two cases for the image to be twice the size of the object:

Case 1: Real and inverted image For a real and inverted image, the magnification $m = -2$. Substitute $m = -2$ and $f = 15 \text{ cm}$ into the formula for $u$: $u_1 = \frac{15(1 - (-2))}{-2}$ $u_1 = \frac{15(1 + 2)}{-2}$ $u_1 = \frac{15 \times 3}{-2}$ $u_1 = \frac{45}{-2}$ $u_1 = -22.5 \text{ cm}$ The object is placed 22.5 cm in front of the lens.

Case 2: Virtual and erect image For a virtual and erect image, the magnification $m = +2$. Substitute $m = +2$ and $f = 15 \text{ cm}$ into the formula for $u$: $u_2 = \frac{15(1 - 2)}{2}$ $u_2 = \frac{15(-1)}{2}$ $u_2 = \frac{-15}{2}$ $u_2 = -7.5 \text{ cm}$ The object is placed 7.5 cm in front of the lens.

The difference between the two object positions is: Difference $= |u_1 - u_2|$ Difference $= |-22.5 \text{ cm} - (-7.5 \text{ cm})|$ Difference $= |-22.5 \text{ cm} + 7.5 \text{ cm}|$ Difference $= |-15 \text{ cm}|$ Difference $= 15 \text{ cm}$

The calculated difference matches the value given in the question.

The final answer is 15.

Explanation of the solution: The magnification $m$ can be $\pm 2$. For $m=-2$ (real image), the object distance $u_1 = \frac{f(1-m)}{m} = \frac{15(1-(-2))}{-2} = -22.5 \text{ cm}$. For $m=+2$ (virtual image), the object distance $u_2 = \frac{f(1-2)}{2} = -7.5 \text{ cm}$. The difference between these two object positions is $|-22.5 - (-7.5)| = |-15| = 15 \text{ cm}$.