Question

A screen is placed a distance d = 43.0 cm to the right of a small...

A screen is placed a distance d = 43.0 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +5.10 cm be placed if the real image of the object is at the location of the screen? (Hint: Use the thin lens equation in the form s' = sf/s-f , so s'(s − f) = sf. Write s' = d − s and solve for s in the resulting quadratic equation.)

Find the shorter distance and the longer distance (these values are 5.913146565 and 37.08685343)

For this lens, what is the smallest value d can have and an image be formed on the screen?

The second half is the part I don't understand

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Dear student,

Find this solution, and RATE IT ,If you find it is helpful .your rating is very important to me.If any incorrectness ,kindly let me know I will rectify them soon.

Thanks for asking ..

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
An object is placed 53.5 cm from a screen. (a) Where should a converging lens of...
An object is placed 53.5 cm from a screen. (a) Where should a converging lens of focal length 9.0 cm be placed to form an image on the screen? shorter distance_________cm from the screen farther distance___________cm from the screen (b) Find the magnification of the lens. magnification if placed at the shorter distance ________ magnification if placed at the farther distance__________
A converging lens of focal length f1 = +22.5 cm is placed at a distance d...
A converging lens of focal length f1 = +22.5 cm is placed at a distance d = 60.0 cm to the left of a diverging lens of focal length f2 = −30.0 cm. An object is placed on the common optical axis of the two lenses with its base 45.0 cm to the left of the converging lens. (The thin-lens approximation may be assumed to hold.) (a) Calculate the location of the final image and its overall magnification with respect...
A converging lens of focal length f is placed between an object and a screen located...
A converging lens of focal length f is placed between an object and a screen located a distance d to the right of the object. Given that 4f < d, find the largest of two possible locations in cm for the lens as measured from the object so that a sharp image is formed on the screen. (Input your answer in 2 significant figures without unit)
A real, upright object of 3.50 cm in height is placed to the left of a...
A real, upright object of 3.50 cm in height is placed to the left of a converging lens with focal length 12.0 cm. A diverging lens with focal length 6.00 cm is located 20.0 cm to the right of the converging lens. Read carefully: The final image formed by the two lens system is 5.00 cm to the right of the converging lens. Draw a simple yet clear diagram of the setup, labeling lenses, object distances, and images distances as...
A converging lens is placed at x = 0, a distance d = 11.0 cm to...
A converging lens is placed at x = 0, a distance d = 11.0 cm to the left of a diverging lens as in the figure below (where FC and FD locate the focal points for the converging and the diverging lens, respectively). An object is located at x = −1.10 cm to the left of the converging lens and the focal lengths of the converging and diverging lenses are 3.50 cm and −8.60 cm, respectively. HINT (a) Determine the...
A converging lens is placed at x = 0, a distance d = 11.0 cm to...
A converging lens is placed at x = 0, a distance d = 11.0 cm to the left of a diverging lens as in the figure below (where FC and FD locate the focal points for the converging and the diverging lens, respectively). An object is located at x = −1.10 cm to the left of the converging lens and the focal lengths of the converging and diverging lenses are 3.50 cm and −8.60 cm, respectively. HINT (a) Determine the...
For a particular converging lens, an object a certain distance to the left of the lens...
For a particular converging lens, an object a certain distance to the left of the lens produces a clear image on a a screen a distance 30.2 cm to the right of the lens. When a second lens is placed 15.4 cm to the right of the first lens, the screen has to be moved 18.7 cm to the right of its previous location to produce a clear image. Determine the focal length of the second lens.
When an object is placed at the proper distance to the left of a converging lens,...
When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30cm to the right of the lens. A diverging lens with focal length f=-24cm is now placed 15cm to the right of the converging lens. How far and in which direction do we need to move the screen to get a sharp image? How much larger is the new image?
A bright object and a viewing screen are separated by a distance of 87.6 cm ....
A bright object and a viewing screen are separated by a distance of 87.6 cm . Part A At what location(s) between the object and the screen should a lens of focal length 18.5 cm be placed in order to produce a sharp image on the screen? [Hint: first draw a diagram.] If there is more than one answer, separate them by a comma.
A converging lens and a concave mirror are placed a distance D = 133 cm apart....
A converging lens and a concave mirror are placed a distance D = 133 cm apart. The lens is placed such that it is to the left of the refelcting side of the mirror. The lens has focal length f1 = 5 cm, and the mirror has focal length f2 = 6 cm. If an object is placed a distance of do = 35 cm to the left of the lens, what is the image distance in centimeters from the...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT