t. An ojoce is eago be ferned y ths pain otiye lens The recal longs otke s 10 om
ID: 2032415 • Letter: T
Question
t. An ojoce is eago be ferned y ths pain otiye lens The recal longs otke s 10 omhll he positives Is the image rral of imaginery? RolJCro'? Mark the position of this image on the figure 5. There is negative lst edia negative lení to the right orte puskive len, at a distance of-10 cm amny, what the distance between ees this aegative lers and the image fomed by the positive lens? 6. The image formed by the positive lees is the object of the negative lens. Is the image formed by positive lens to the right or the iefl of the negative lens? In this case, is the object distance positive or negative? To the negative lens, 7. The focal length of the negative lens is also 10 em (negative). Calculase where the image will be formed by the negative lens mark the position of this image on the figure. Is the image real or virtual? For a diverging lens, the focal length is negalive by definition. Thes in the thin lens equation, If the object distance is positive, then the image distance will swnys be negative. How did we make the diverging lens to form a real image? 0-3P.2 December 19, 2017Explanation / Answer
4) f1 = 10 cm
u1 = 30 cm
let v is the image distance.
use, 1/u1 + 1/v1 = 1/f1
1/v1 = 1/f1 - 1/u1
1/v1 = 1/10 - 1/30
v1 = 15 cm
The image is real.
5) u2 = v2 - L
= 15 - 10
= 5 cm
6) object distance is negative. because, object is right side to the lens.
so, u2 = -5 cm
7) f2 = -10 cm
u2 = -5 cm
let v2 is the image distance from the negative lens.
use, 1/u2 + 1/v2 = 1/f2
1/v2 = 1/f2 - 1/u2
1/v2 = 1/(-10) - 1/(-5)
v2 = 10 cm
The image is real .
When the image of first lens is right to the diverging lens and in between the focal point and lens of diverging lens, the image acts as virtual object for the diverging lens.
and we can get a real image.
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