A system of two converging lenses forms an image of an arrow as shown. The first

Question

A system of two converging lenses forms an image of an arrow as shown. The first lens is located at x = 0 and has a focal length of f1 = 10.6 cm. The second lens is located at x = x2 = 57.5 cm and has a focal length of f2 = 18.4 cm. The tip of the object arrow is located at (x,y) = (xo,yo) = (-15.9 cm, 6.2 cm).

1)

What is x1, the x co-ordinate of image of the arrow formed by the first lens?

_____-cm

2)

What is y1, the y co-ordinate of the image of the tip of the arrow formed by the first lens?

_____cm

3)

What is x3, the x co-ordinate of image of the arrow formed by the two lens system?

____cm

4)

What is y3, the y co-ordinate of the image of the tip of the arrow formed by the two lens system?

___cm

5)

What is the nature of the final image relative to the object?

a. Real and Inverted

b. Real and Upright

c. Virtual and Inverted

d. Virtual and Upright

6)

Which of the following changes to the locations of the lenses would result in a virtual and inverted image of the original object arrow?

a. Move the first lens to x = -10.6 cm, keeping the second lens at x = 57.5 cm.

b. Move the second lens to x = 66.7 cm, keeping the first lens at x = 0.

c. Move the second lens to x = 41 cm, keeping the first lens at x = 0.

d. None of the above moves will produce a virtual inverted image of the object arrow.

Help...

Two Converging Lenses 1 23 45 6 A system of two converging lenses forms an image of an arrow as shown. The first lens is located at x 0 and has a focal length of f- 10.6 cm. The second lens is located at xx2-57.5 cm and has a focal length of f2 - 18.4 cm. The tip of the object arrow is located at (x.y) = (Xo.yo) = (-15.9 cm, 6.2 cm) (oy) 1) What is x1, the x co-ordinate of image of the arrow formed by the first lens? Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 2) What is y1, the y co-ordinate of the image of the tip of the arrow formed by the first lens? cm Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 3) What is x3, the x co-ordinate of image of the arrow formed by the two lens system? Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 4) What is ys, the y co-ordinate of the image of the tip of the arrow formed by the two lens system? cm Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 5) What is the nature of the final image relative to the object? Real and Inverted O Real and Upright O Virtual and Inverted O Virtual and Upright Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 6) Which of the following changes to the locations of the lenses would result in a virtual and inverted image of the original object arrow! O Move the first lens to x =-10.6 cm, keeping the second lens at x = 57.5 cm O Move the second lens to x = 66.7 cm, keeping the first lens at x = 0 O Move the second lens to x = 41 cm, keeping the first lens at x = 0 O None of the above moves will produce a virtual inverted image of the object arrow Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question.

Explanation / Answer

(1)

F=10.6 cm

distance of object from lense u =15.9 cm , Let distance of image from lense = v

we have relation

1/v + 1/u = 1/f

1/v +1/(15.9) = 1/(10.6)

1/v=0.031

v=32.25 Cm ANS

(2)

We also have relation

hi / ho = - v/u

We have hi =6.2 cm

6.2 / ho=-(32.25) / (15.9)

ho=- 3.06 cm ANS

(c)

Image formed by lense 1 is at =32.25 cm and lense is at 57.5 cm

so Here u=( 57.5-32.25)=25.25 cm

f=18.4 cm

again using

1/v + 1/u = 1/f

1/v + 1/(25.25)=1/(18.4)

1/v=0.015

v=66.67 cm

(d)

hi / ho = - v/u

Here hi=- 3.06

(- 3.06) / h0= - (66.67) / (25.25)

ho=1.16 cm

(e)

Real and upright

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