Prove that if an object is at a distance of twice the focal length from a lens the image is the same size as the object. Show this using an algebraic method
Here ,
let the focal length is f
object distance = 2 * focal length
object distance , do = 2 * f
Using the lens formula
1/f = 1/di + 1/do
1/f = 1/2f + 1/di
di = 2f
now , magnfication = -di/do
magnfication = -2f/2f
magnfication = -1
as magnfication is -1
hence , the object height is same as the image height
Prove that if an object is at a distance of twice the focal length from a...
(a) The relation between the object distance S, the image distance S' and the focal length f of a lens producing a real image is given as 1 1 1 S S This is called the Gaussian form of the thin lens formula. Another form of the formula, the Newtonian form, is obtained by considering the distance x, from the object to the first focal point and the distance x; from the second focal point to the image. Show that...
An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image? A:+2/3 B:-3/2 C:+3/2 D:-2/3 E:+1/3 F:-1/3 -An explanation would be appreciated
1. A 4.00-cm tall object is placed a distance of 48 cm from a concave mirror having a focal length of 16cm. Determine the image distance and the image size. 2. A 4.00-cm tall object is placed a distance of 8 cm from a concave mirror having a focal length of 16cm. Determine the image distance and the image size. 3. Determine the image distance and image height for a 5.00-cm tall object. placed 30.0 cm Infront of from a convex mirror...
For a converging lens with focal length of 15 cm, an object at twice the focal length. Calculate the magnification of the image (Note: Negative means the image is inverted.) A. + 1 B. -1 C. +2 D. -2
An object is located 26.0 cm from a certain lens. The lens forms a real image that is twice as high as the object. What is the focal length of this lens? Now replace the lens used in Part A with another lens. The new lens is a diverging lens whose focal points are at the same distance from the lens as the focal points of the first lens. If the object is 5.00 cm high, what is the height...
1. Using the thin lens equation (1/f=(1/s)+(1/s')), prove that the focal length of a lens is equal to the image distance when there is parallel incident light 2. There is a specific case where for a single converging lens the object distance equals the image distance. Find the relation between the focal length and the object distance in this case.
Measured: Focal length = Object distance = Object height = Measured from ray diagram: Image distance = Image height = Calculate magnification from data: Image characteristics: From measured focal length and object distance, calculate the expected imare distance from the mirror equation. Compare the calculated and the measured image distance by calculating the percent error between them. Page 2 of 4
You need to use a 22-cm-focal-length lens to produce an inverted image twice the height of an object. 1. At what distance from the object should the lens be placed? Express your answer to two significant figures and include the appropriate units.
6. A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size. 7. A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. (NOTE: this is the same object and the same lens as in #6, only this time the object is placed closer to...
A convergent lens has a focal length of 10.8 cm. The object distance is 19.6 cm. Find the distance of the image from the center of the lens. Answer in units of cm. Find the magnification.