Question

Problem Statement A thin converging lens is found to form an image of a distant building 24 cm from the lens. If an insect is now placed 16 cm from this lens, how far FROM THE INSECT will its image be formed? Visual Representation Draw an optical axis with the lens centered on the axis. Represent the object at the correct distance from the lens Draw the three "special rays" from the top of the object • Extend the rays until they converge. Draw the image Mathematical Representation Start with the thin lens equation and definition of magnification • Substitute in the known values Solve for the desired unknown Check units and assess if your answer is reasonable

Please complete the visual portion as well

Answer 1

Visual Representation part:

As building is too far, we can assume it at infinite. And we know that rays from infinite, converge at focus point or focal plane. So, we can say that the focal length of thin converging lens is 24cm. The ray diagram is as below-

object image TI ALL си V CS Scanned with CamScanner

Now, if insect is placed at 16 cm, then the ray diagram will be as below-

B Ti A F image Pobject 0 L U CS Scanned with CamScanner

Hence, the insect will be 16 cm before the lens, i.e

и = -16cm

And

= 24cm

So,

1 = -48cm

Hence, From the insect, the image will be at a distance of (48-16)cm=32cm

Mathematical Representation:

If is distance of object, is the distance of image from point and is the focal length of thin convex lens, then the Lens equation is as below-

  $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ here, sign of and is already considered.

Magnification: Magnification is an unitless quantity which gives the ratio of the size of the image with respect to the size of object.

i.e Magnification, $M=\frac{I}{O}=\frac{v}{u}$

Calculations:

Given that $u=16cm;f=24cm$

So, Substituting in Lens formula,

$\frac{1}{24}=\frac{1}{16}+\frac{1}{v}$

So, $\frac{1}{v}=\frac{1}{24}-\frac{1}{16}=\frac{2-3}{48}=-\frac{1}{48}$

Hence,    (sign of v denotes the position of v, left or right to point o)

1 = -48cm

magnitude of $v=48cm$

Now, Magnification

$M=\frac{v}{u}=\frac{48}{16}=3$

So, According to our mathematical calculations, the image is 3 times larger than the object(insect) and is virtual(same side of object) and lie behind the object. It's also verified in our ray diagram.