12:24 PM iPad 22% ust49wa.theexpertta.com Home Student: Megan Montgomery@wsu.edu

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12:24 PM iPad 22% ust49wa.theexpertta.com Home Student: Megan Montgomery@wsu.edu My Account Log Out Class Management Help Homework #4: Optical Devices Begin Date: 2/17/2017 1:59:00 PM Due Date: 2/23/2017 11:59:00 PM End Date: 5/8/2017 11:59:00 PM (13%) Problem i: People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. i A 33% Part (a) What is the power in Dof the eyes of a woman who can see an object clearly at a distance of only 9.5 Assume the distance to her retina from the lens in her eye is 2.00 cm. i A 33% Part (b) What is the size in mm of an image of a 1.25 mm object, such as lettering inside a ring, held at this distance A 33% Part (c) What would the size in mm of the image be if the object were held at the normal 25 cm distance Grade Summary Deductions 0% Potential 100 Submissions tan Attempts remaining: 20 cotan0 asino acos per attempt) (0% atano acotan0 sinh() detailed view cosh() tanh0 cotanh0 O Degrees O Radians Hint Submit I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. All content 2017 Expert TA, LLC

Explanation / Answer

part a:

let focal length be f.

object distance =u=-9.5 cm

image distance=v=2 cm

then (1/v)-(1/u)=1/f

==>1/f=(1/2)+(1/9.5)

==>f=1.6522 cm

then power=1/f

=1/(1.6522*0.01) D

=60.526 D

part b:

iamge size/object size=image distance/object distance

==>image size/1.25 mm=2/9.5

==>image size=(2/9.5)*1.25 mm=0.26316 mm

part c:

u=-25 cm

f=1.6522 cm

then if image distance is v

(1/v)+(1/25)=1/1.6522

==>v=1.7691 mm

then size of image=1.25*(1.7691/25) mm
=0.088455 mm

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