I did my work and i felt like I\'m missing something. I want to compare my answe

Question

I did my work and i felt like I'm missing something. I want to compare my answer with yours. Please show me your steps so I can chat with you and understand. Please remember Chegg's 5 subsets per question policy, which means you should do all of 1 and 2.
For multiple choice questions, circle the letter of the Show your work on free-response questions. Be sure to use proper units and significant figares in your final answers Organization and neatness count, too. Do NOT use red ink- please reserve that color for grading one best answer (unless imore than one answer is asked for 1.You do NOT neel to show your work for this question. Be sure to include correct units. ± signs, and sig. Jig. a. You place a 300-cm-tall object 900 cm avay from a comerging lens with a 400 cm focal length. Usingour standard plusimines sign conventions, find the following values 4 pts) What is the position of the image? What is the height of the image ? What is the magnification of the image? Is the image real or virtual b. You place the same 30.0-cm-tall object 90.0 cm away from a diverging lens with a 400 cm focal length What is the height of the image? 4 pts.) What is the position of the image? Is the image real or virtual? What is the magnification of the image? 2. (4 pts) A spherical, concave mirror has a radius of curvature of 85 cm. You have a small, (brighty lit) plastic action figure that is 10.0 cm tall, and you want to create a real image of it that is 1 5 times as large as the original object. (Hint: Watch your signs- will the real image be upright or inverted?) At what distance from the mirror should you place the object? Show your work completely

Explanation / Answer

1. a. h = 30 cm

object distance u = - 90 cm

convergin lens focal length, f = 40 cm

image position = v

using lens formula

1/v - 1/u = 1/f

1/v = 1/40 - 1/90

v = 72 cm

image height, h' = h*(v/u) = -24 cm

magnification = h'/h = -0.8

for -ve magnification, the image is real

b. f = -40 cm

1/v = -1/40 - 1/90

v = -27.6923 cm

image height = h*(v/u) = 9.2307 cm

magnification = 0.3076

the image is virtual as magnificatino is _ve

2. concave mirror, R = 85 cm

h = 10 cm

h' = -15 cm ( real image)

let the object distance be u

then oimage distance = v

-v/u = -1.5

v = 1.5u

hence

1/v + 1/u = 1/f= -2/R

1/1.5 + 1 = -2u/85

u = -70.833 cm

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