To find the image distance for varying object distances in case of a convex lens of focal length 15 cm, a student obtains on a screen a sharp image of a bright object by placing it at 20 cm distance from the lens. After that he gradually moves the object away from the lens and each time focuses the image on the screen.(a) In which direction-towards or away from the lens does he move the screen to focus the object?(b) How does the size of image change?(c) Approximately at what distance does he obtain the image of magnication –1?(d) How does the intensity of image change as the object moves farther and farther away from the lens?
Answers
if any mistake is their please tell me in comments
The lens formula is given as :
1/f = 1/u + 1/v
f = The focal length
u = the image distance
v = the object distance.
f = 15 cm
v = 20 cm
The image distance is given as :
1/15 - 1/20 = 1/60
The image distance is given as :
u = 60 cm
Taking the object further by 10 cm we have :
v = 20 + 10 = 30
1/15 - 1/30 = 1/30
u = 30 cm
Reducing the image distance by 10 we have :
v = 20 -10 = 10 cm
1/15 - 1/10 = -1/30
u = -30 cm
From the calculations above we see that as we move the object closer to the lens the image distance becomes negative which implies that the image is virtual.
a) To focus the object we therefore needs to move the object away from the lens since the image distance remains positive.
b) Magnification = image distance / object distance
= 60/20 = 3
The image becomes three times the object.
c) Magnification = image distance/object distance.
-1 = di/20
-1 × 20 = -20
= -20 cm
d) The image becomes more clear as we move the object further away.