Question

Case 1. The relationship between the distance of object from the lens(u), distance of image from the lens(v) and the focal length(t) of the lens is called lens formula it can be written as1/4-1/-1/u. The size of image formed by a lens depends on the position of the object from the lens. A lens of short focal length has more power whereas a lens of long focal length has less power. When the lens is convex, the power is positive and for concave lens, the power is negative. The magnification produced by a lens is the ratio of height of image to the height of object as the size of the image relative to the object is given by linear magnification(m) When mis negative, image formed is real and when mis positive, Image formed is virtual. If m 1, size of image is smaller than the objectif m>1, size of image is larger than the object.

A) an obj 4cm on height is placed at a distance of 10 cm from a convex lens of focal length 20 cm The position of image is ?

B) in above question size of image is?

C) an obj is placed 50cm from a concave lens and produces a virtual image at a distance of 10cm on front of lens. The focal length of lens is ?

D) A concave lens of focal length 5cm , the power of lens is ?​

Answer 1

Answer:

I only Know 1 Answer

(A)

u=−10cm 

h1=4cm 

f=20cm 

v1−u1=f1 

v1−−101=201 

v1=201−101=−201 

v=−20cm (Image is 20cm in front of the convex lens)

 

m=uv=−1020=−2 

m=h1h2=−2 

4h2=−2   

h2=−8cm 

Image is 8cm in size and is real and inverted . 

Answer 2

All question parts make use of the lens formula which is:

                                   $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

where v is the position of the image

           u is the position of the object

           f is the focal length of the lens

Now, we can solve all the parts.

A) Given: Focal length of convex lens = +20 cm

                Object distance = - 10 cm

                Height of object = 4 cm

To Find: Position of image (v)

Solution:

• Applying lens formula and substituting given values :

                       $\frac{1}{v}$ - (-$\frac{1}{10}$) = $\frac{1}{20}$

                   ⇒ $\frac{1}{v}$ = $\frac{1}{20}$ - $\frac{1}{10}$

                  ⇒  $\frac{1}{v}$ = $\frac{-1}{20}$

                  ⇒ v = -20 cm

Therefore, the image is formed 20 cm behind the lens on the focal point.

B) Given : Same as in part (A)

   To Find: Size of image

Solution:

• Magnification produced by lens is given by

                                      m = $\frac{h_{i} }{h_{o} }$ = $\frac{v}{u}$              

where, $h_{i}$ is height of image

           $h_{o}$ is height of object

• Substituting known values:

                         m = $\frac{h_{i} }{4}$ = $\frac{-20}{-10}$

                     ⇒ $h_{i}$ = 2 × 4 =8 cm

Therefore, size of image is 8 cm.

C) Given: Object distance = -50 cm

                Image distance = - 10 cm

To Find: Focal length of concave lens

Solution:

• Applying lens formula substituting given values :

                   ⇒  $\frac{-1}{10}$ - (-$\frac{1}{50}$) = $\frac{1}{f}$

                  ⇒  $\frac{1}{f}$ = $\frac{-4}{50}$

                 ⇒ f = -12.5 cm

Therefore, the focal length of the concave lens is -12.5 cm.

D) Given: Concave lens of focal length = -5 cm

    To Find: Power of given concave lens

Solution:

• The Power of the lens is defined as the reciprocal of focal length.
• Its unit is Dioptres (D).
• Mathematically,

                 P = $\frac{100}{f}$ if focal length is in cm

Substituting known values in formula:

        P = $\frac{100}{-5}$ = -20 D

Therefore, the power of the concave lens is -20 D.