How is an image projected on the retina of your eye? Describe this by using L.O.S.T (Lost explained below)
L:- Location of the image (Further or closer to the lens or mirror)
O:- Orientation of the image (Upright or inverted)
S:- Size of the image (smaller, larger)
T: Type of image (real or virtual)
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The human eye pupil can be regarded as a converging lens having variable focal length. The focal distance for this lens when the object placed is at infinity is equal to the distance from pupil to retina, which is about 1.7 cm. Because the image need to be in the same place (on the retina) this focal distance becomes smaller for closer objects.
The figure below presents the formation of an object image into a converging lens (like the eye pupil). The equation for a converging lens is
`1/(x1) +1/(x2) =1/f` where
x1 is positive if the object is to the left of the lens
x2 is positive if the image is to the right of the converging lens
f is positive for convergent lenses.
1. Location of the image
By taking an object at x1=infinity (to the left of the eye) and f = 1.7 cm we get
`1/(x2) =(1/f) -(1/(x1)) =1/f` , hence `x2 =f =1.7 cm` to the right of the lens.
By taking an object at x1=2 m in front of the eye (to the left) and f=1.685 cm we get
`1/(x2) =(1/f) -(1/(x1)) =(1/0.01685) -(1/2)` , hence
`X2 =0.017 m=1.7 cm` to the right of the lens.
2. Orientation of the image
The magnification of a converging lens is by definition
`M = -(Y2)/(Y1) = -(X2)/(X1)`
Since both X1 and X2 are positive it implies `M < 0`
Thus, the image is inverted with reference to the object position.
3. Size of image
Since always `X2 =0.017 m =1.7 cm`
and `X1 cong 0.1-:100 m` then the absolute value of magnification
`Abs(M) < 1` . Thus the image is always smaller than the object.
4. Type of image
Since `X2 > 0` (to the right of the eye) and the image is formed at the intersection of real light rays , the image is always real.
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