iPad us tus 12:24 PM 22% ust49wa.theexpertta.com Home I Student: Megan Montgomer

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iPad us tus 12:24 PM 22% ust49wa.theexpertta.com Home I Student: Megan Montgomery@wsu.edu My Account Log Out Class Management I 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 5: An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. A 20% Part (a) What is the image distance (in cm) for this configuration? A 20% Part (b) What is this image's magnification? A 20% Part cm from the objective. What is the image distance for the eyepiece in (c) An eyepiece with a 2.1 cm focal length is placed 19, 76 cm i A 20% Part (d) What magnification is produced by the eyepiece A 20% Part (e) What is the overall magnification? Grade Summary Deductions Potential 100 09% sino Submissions 7 8 9 Attempts remaining: 20 cotano asino acos0 (0% per attempt) ata no acotano sinh0 detailed view cosh0 tanh0 Cota O Degrees O Radians Submit Hint I give up! Hints deduction per hint. Hints remaining 2 Feedback 0% deduction per feedback. All content 2017 Expert TA, LLC

Explanation / Answer

Given that

The focal length of the lens (f) =0.300cm

The object distance is (u) =0.305cm

The image distance is given by (v) =?

We know that

1/f =1/u+1/v

1/v =1/f-1/u =1/0.300-1/0.305 =3.333-3.278=0.055

v =1/0.055

Then the image distanc is v=18.18cm image distance for the first lens

b)

The image magnification is given by

m1 =-v/u =-18.18/0.305 =-59.606cm

c)

Given that

The fical length of the eye piece is (f) =2.1cm

The objective distance is (u) =19.76cm-18.18cm =1.58cm

The image distance is given by

1/f =1/u+1/v

1/v =1/f-1/u =1/2.1-1/1.58 =0.476-0.632=-0.1569

v =1/-0.1569=-6.373cm image distance from the second lens

d)

The magnification produced by the eyepiece is

m2 =-v/u =-(-6.373)/1.58 =4.033

e)

The overall magnification is given by

M =m1*m2 = -(59.606)*(4.033) =-240.423

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