Ảnh của thấu kính và gương có dấu
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| virtual image/ˈvɜːtʃuːəl ˈɪmɪdʒ/ | ảnh ảo |
| converging/kənˈvɜːdʒɪŋ/ | Bringing parallel rays together to a focus. |
| magnification/ˌmæɡnɪfɪˈkeɪʃn/ | The ratio of image height to object height, negative for an inverted image. |
A negative answer is a location
- Solving the lens equation gives $s' = -20$ cm. That is not an error to fix with an absolute value: it says the image is virtual, on the incoming side.
- GRE geometric optics tests signed lens and mirror equations, image character and two-lens systems.
The sign convention
- Real-is-positive: real object $s > 0$, real image $s' > 0$, virtual image 虚像 $s' < 0$. $\dfrac1f = \dfrac1s + \dfrac1{s'}$, $M = -\dfrac{s'}{s}$.
- Converging 会聚 lens $f > 0$; diverging $f < 0$. Concave mirror $f = R/2 > 0$; convex $f < 0$.
A negative image distance should be replaced by its absolute value before interpreting it.
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The sign marks a virtual image.
Lens images
- $s > f$: real, inverted. $s < f$: virtual, upright, enlarged (magnifier). $s = f$: parallel rays, no image. Diverging lens, real object: virtual, upright, smaller.
- Rays: parallel → through the far focus; through the centre → undeviated.
Converging lens $f = 20$ cm, object 60 cm away. Image distance in cm?
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The lens equation $1/s' = 1/f - 1/s = 1/20 - 1/60 = 1/30$ gives a real image 30 cm away.
Same lens, object 10 cm away. Magnification?
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With $s = 10$ cm inside the focal length, $1/s' = 1/20 - 1/10$, so $s' = -20$ cm (virtual) and $M = -s'/s = 2$.
Diverging lens $f = -15$ cm, object 30 cm away. Image distance in cm?
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$1/s' = 1/f - 1/s = -1/15 - 1/30 = -1/10$, so the image is virtual, 10 cm in front of the lens: $s' = -10$ cm.
Mirrors
- Same equation and signs. A real mirror image sits in front of the mirror; virtual behind it.
- Convex mirror $f = -12$ cm, $s = 24$ cm: $\tfrac1{s'} = -\tfrac1{12} - \tfrac1{24}$, $s' = -8$ cm, $M = +\tfrac13$: upright, behind the mirror.
What kind of image?
Predict image character from element and object position.
A concave mirror with $R = 40$ cm and an object 30 cm away gives an image
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
f = 20: 1/s′ = 1/20 − 1/30 = 1/60; M = −2.
Two lenses
- Find the first image, then measure the object distance for the second lens from its position; converging-in rays make a virtual object ($s < 0$). The total magnification 放大率 is $M = M_1M_2$.
$f = 10$ cm lens at $x = 0$, object at $x = -30$: image at $x = 15$, $M_1 = -\tfrac12$. Second $f = 10$ cm lens at $x = 40$: $s_2 = 25$, $s_2' = \tfrac{50}{3}$ cm, $M_2 = -\tfrac23$. Total $M = +\tfrac13$.
The spacing between two lenses is not automatically the second object distance: measure from the first image to the second lens, with its sign.
Recap
- $1/f = 1/s + 1/s'$, $M = -s'/s$, signs kept to the end.
- Negative $s'$: virtual image; positive $M$: upright.
- In a system, each lens takes the previous image as its object.
- converging: Bringing parallel rays together to a focus.
- magnification: The ratio of image height to object height, negative for an inverted image.