Question

A converging lens with a focal length of 9.30 cm forms an image of a 5.60-mm-tall real object that is to the left of the lens. the image is 1.90 cm tall and erect

where is the object located? follow the sign rules. express your answer with the appropriate units.

Answer 1

The object is located approximately 2.53 cm to the left of the converging lens with a focal length of 9.30 cm, forming an erect image 1.90 cm tall on the right side.

To find the object's location, you can use the lens formula:

`\[ (1)/(f) = (1)/(d_o) + (1)/(d_i) \]`

where:

`- \( f \)`is the focal length of the lens,

-
`\( d_o \)` is the object distance (distance from the lens to the object),

`- \( d_i \)`is the image distance (distance from the lens to the image).

The sign conventions are as follows:

-
`\( f \)` is positive for converging lenses,

-
`\( d_o \)` is positive for objects on the same side as the incident light (left side for a converging lens),

-
`\( d_i \)` is positive for images formed on the opposite side of the lens from the incident light.

Given:

`- \( f = 9.30 \, \text{cm} \) (positive for a converging lens),`

`- \( h_o = 5.60 \, \text{mm} \) (object height),`

`- \( h_i = 1.90 \, \text{cm} \) (image height).`

First, convert all measurements to the same unit (centimeters):

`- \( h_o = 0.56 \, \text{cm} \) (5.60 mm converted to cm).`

Now, use the magnification formula
`\( (h_i)/(h_o) = -(d_i)/(d_o) \)`to find
`\( d_i \)`, where the negative sign indicates an inverted image for the converging lens:

`\[ (h_i)/(h_o) = -(d_i)/(d_o) \]`

`\[ (1.90)/(0.56) = -(d_i)/(d_o) \]`

`\[ d_i = -(1.90)/(0.56) * d_o \]`

Now, substitute this expression for
`\( d_i \)` into the lens formula:

`\[ (1)/(f) = (1)/(d_o) - (0.56)/(1.90) \]`

Solve for
`\( d_o \):`

`\[ (1)/(9.30) = (1)/(d_o) - (0.56)/(1.90) \]`

`\[ (1)/(d_o) = (1)/(9.30) + (0.56)/(1.90) \]`

`\[ (1)/(d_o) = (0.1)/(1) + (0.56)/(1.90) \]`

`\[ (1)/(d_o) \approx (0.1 * 1.90 + 0.56 * 1)/(1.90) \]`

`\[ (1)/(d_o) \approx (0.19 + 0.56)/(1.90) \]`

`\[ (1)/(d_o) \approx (0.75)/(1.90) \]`

Now, find
`\( d_o \):`

`\[ d_o \approx (1.90)/(0.75) \]`

`\[ d_o \approx 2.53 \, \text{cm} \]`

So, the object is located approximately
`\( 2.53 \, \text{cm} \)` to the left of the converging lens.