Test: Lense Formula - Class 10 MCQ

Magnification = the height of the image ÷ by the height of the object 
  M=36/9 =4

We know, height of the object is 4 cm h1, height of the image is 3 cm, h2. 
So we have, m = h2/h1 
=> m = 3/4  
=> m = 0.75
Therefore, magnification of the lens is 0.75 

72.f = - 0.2 m, v = + 0.3 mThe lens formula 1v - 1u = 1f
 1u = 1v - 1f = 10.3 + 10.2
   = 0.50/006 
u = 0.12 m

Since, it is a diverging lens, that is, a concave lens, we have
u = -10 cm, v = -6.5 cm

1/v - 1/u 
1/f   =-1/u - 1/u
1/f   = - 2/u
 u=-2f
 u=-24 cm

If the object is placed at distance of f/2 in front of convex lens of focal length 'f'.
If the object is at f/2 distance then it means it is between the focus and optical centre of lens. And we know, that when an object is placed between the focus and optical centre of convex lens then the image is formed behind the object and the nature of image is virtual, errect and magnified. 
By lens formula ; 
1/f = 1/v - 1/u 
1/f  = 1/v - 1/ (-f/2) 
Solving this we get, 
1/ v = -f/2 /  f²/2 
v = f² /2 / -f/2 = -f 

Focal length= -20(as it is concave lens)
v= -15 (as concave lens always forms virtual and erect image on left of lens)
Putting these values in lens formula,
1/ -20 - 1/u = 1/ -15
-1/ u= 1/-15 + 1/20
-1/u = -4+3/60
-1/u = -1/60
-u = -60
[u =60]

An object should be placed between F₁ and 2 F₁ so as to obtain the image formation used in slide projector. This is because, the image has to be enlarged and has to form at a little larger distance from the projector.

Given : f = 10 cm , m = 2
v / u = 2
v = 2u 
v = 2u -------(1)
according to lens formula , 
1/v - 1/u = 1/f
1/v - 1/u = 1/10 --------(2)
substitute (1) in (2)
1/2u - 1/u = 1/10
1 - 2 / 2u = 1/10
-1 / 2u = 1/10
-10 = 2u
u = -10/2
u = -5 cm
Therfore object should be placed 5 cm away from the lens.