a) How far is the image from the lens? b) What is the size of the image?[ c) Is it inverted (yes or no)? 1 d) Is it real or virtual? [-] e) What is the magnification of the image?

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Chapter1: Units, Trigonometry. And Vectors
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a) How far is the image from the lens?
b) What is the size of the image?[1
c) Is it inverted (yes or no)? 1
d) Is it real or virtual? [-]
e) What is the magnification of the image?
Transcribed Image Text:a) How far is the image from the lens? b) What is the size of the image?[1 c) Is it inverted (yes or no)? 1 d) Is it real or virtual? [-] e) What is the magnification of the image?
11: Use ray tracing (or any other correct method) to find the image of the arrow formed by the thin
converging lens. One focal point is shown, 20 cm to the left of the lens. (1 square is 2 cm wide.) The
arrow is 7.5 cm to the left of the lens.
f
Transcribed Image Text:11: Use ray tracing (or any other correct method) to find the image of the arrow formed by the thin converging lens. One focal point is shown, 20 cm to the left of the lens. (1 square is 2 cm wide.) The arrow is 7.5 cm to the left of the lens. f
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To calculate the distance of the image from the lens and the size of the image and to describe whether the image is inverted or not.

Given:
Distance between the lens and the object: u=7.5 cm
Height of the object: ho=4 cm
Focal length of the lens: f=20 cm

To obtain:
a) The distance between the image and the lens: i
b) The size of the image: hi
c) To determine whether the image is inverted or not.

The lens formula in the Cartesian coordinate system is given as:
1o+1f=1i
where,
o is the object distance,
f is the focal length of the lens and
i is the image distance.

In the Cartesian coordinate system of the lens formula any length measured to the left of the lens is negative and any length measured to the right of the lens is positive. The focal length of a converging lens is always positive and the focal length of a diverging lens is always negative.

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