A) a real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. find the image distance, the image height, if the image is real or virtual (explain in detail) and if the image is erect or inverted (explain in detail). b) the real object is now placed 13.0 cm in front of the converging lens. again find the paramaters listed above. graphically show the optic configuration for this second object placement.

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

We need to calculate the height of the image

Using formula of height

Where, h' = height of image

h = height of object

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

Put the value into the formula

We need to calculate the height of the image

The image is virtual and erect.

Hence, This is required solution.

where is the photo and the answer choices?

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j