Question

A 2.0cm tall object is placed 40cm from a diverging lens of focal length 15cm . Find the position and size of the image.

Answer 1

A 2.0cm tall object is placed 40cm from a diverging lens of focal length 15cm . The image is virtual, located 30cm from the lens, and is 0.75cm tall.

Step-by-step explanation:

Utilizing the lens formula (1/f = 1/v - 1/u), where f is the focal length, v is the image distance, and u is the object distance, we can determine the image's position. Substituting the given values (f = 15cm, u = -40cm for a diverging lens), the formula rearranges to 1/v = 1/f + 1/u. Solving for 1/v gives 1/v = 1/15 - 1/40, resulting in v = 30cm. The negative sign for u indicates the object is placed on the opposite side of the lens from the incoming light.

The magnification equation (m = -v/u) can then determine the image's size. Plugging in the values (v = 30cm, u = -40cm) yields m = -30/-40, simplifying to m = 0.75. The negative sign in the magnification equation indicates the image is virtual and upright. Therefore, the image formed by the diverging lens is virtual, located 30cm from the lens, and its size is 0.75 times the object's size.

To find the position and size of the image formed by a diverging lens when a 2.0cm tall object is placed 40cm from it, we'll utilize the lens formula and magnification equation.

The lens formula for a diverging lens is given by:

`\[ (1)/(f) = (1)/(v) - (1)/(u) \]`

Where:

- f is the focal length of the lens.

- v is the image distance (distance of the image from the lens).

- u is the object distance (distance of the object from the lens).

Given:

Focal length of the diverging lens, f = 15 cm

Object distance, u = -40cm (Negative because the object is placed on the opposite side of the incoming light for a diverging lens.)

Let's use the lens formula to find the image distance (v ):

`\[ (1)/(f) = (1)/(v) - (1)/(u) \]`

`\[ (1)/(15) = (1)/(v) + (1)/(40) \]`

`\[ (1)/(v) = (1)/(15) - (1)/(40) \]`

`\[ (1)/(v) = (8 - 3)/(120) \]`

`\[ (1)/(v) = (5)/(120) \]`

`\[ v = (120)/(5) \]`

`\[ v = 24 \, \text{cm} \]`
`\[ v = 24 \, \text{cm} \]`
`\[ v = 24 \, \text{cm} \]`

The negative sign for u indicates the object is placed on the opposite side of the lens from the incoming light. The image distance ( v ) obtained is positive, which means the image formed is on the same side as the object, making it a virtual image.

Now, let's use the magnification formula to determine the size of the image:

The magnification formula is given by:

`\[ m = (-v)/(u) \]`

Substituting the values:

`\[ m = (-24)/(-40) \]`

m = 0.6

The negative sign in the magnification indicates that the image is upright, which is typical for virtual images formed by diverging lenses.

Therefore, the image formed by the diverging lens is a virtual image located at 24 cm from the lens and its size is 0.6 times the size of the object.