A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm.What will be the length of the mirror?
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Answered by
5
o calculate this sort of question, we make use of magnification and mirror equations.
- The mirror and magnification equations, make use of ray diagrams to predict the type, orientation, location and size of image formed from an object.
⇒The mirror equation gives a numerical relationship of the object distance(u), image distance (v) and the focal length. This equation is given below:
1/f = 1/u + 1/v
- this is the equation we will use to work out this question.
⇒ On the other hand, the magnification equation gives a numerical relationship of the ratio between the object distance - image distance as compared to the ratio of the object height (h₁) to the image height (h₂).
The equation is given as follows:
u/v = h₁/h₂
⇒ In the diagram below I have attached the steps in calculating the length of the image in the question above:
orrrrrr
.
Here we are using concave mirror, to solve this problems one should know sign convections for spherical mirror-
1)object distance ( u ) always negative, as object always kept in front of mirror
2) image disitance (v) is positive if formed image is virtual and negative if formed image is real
3) Object height ( h1 ) always positive, as object always kept above principal axis
4) image height ( h2) is negative if image formed below principal axis which is inverted and positive if it formed above prinicipal axis which is lateraly inverted
using this knowledge and using Mirror formula and Magnification we can solve Numerical base on Mirror
...Mirror Formula
.. Magnification
I have attached an image in which step by step solution is given of above example.
hoping helps you any doubt comments below follow me
- The mirror and magnification equations, make use of ray diagrams to predict the type, orientation, location and size of image formed from an object.
⇒The mirror equation gives a numerical relationship of the object distance(u), image distance (v) and the focal length. This equation is given below:
1/f = 1/u + 1/v
- this is the equation we will use to work out this question.
⇒ On the other hand, the magnification equation gives a numerical relationship of the ratio between the object distance - image distance as compared to the ratio of the object height (h₁) to the image height (h₂).
The equation is given as follows:
u/v = h₁/h₂
⇒ In the diagram below I have attached the steps in calculating the length of the image in the question above:
orrrrrr
.
Here we are using concave mirror, to solve this problems one should know sign convections for spherical mirror-
1)object distance ( u ) always negative, as object always kept in front of mirror
2) image disitance (v) is positive if formed image is virtual and negative if formed image is real
3) Object height ( h1 ) always positive, as object always kept above principal axis
4) image height ( h2) is negative if image formed below principal axis which is inverted and positive if it formed above prinicipal axis which is lateraly inverted
using this knowledge and using Mirror formula and Magnification we can solve Numerical base on Mirror
...Mirror Formula
.. Magnification
I have attached an image in which step by step solution is given of above example.
hoping helps you any doubt comments below follow me
Attachments:
shreya5768:
thanks
Answered by
4
hello dear here is ur answer
1/f=1/v+1/u F=-10; U=-20
1/v=1/f-1/u
1/v=-1/10+1/20
1/v=-2+1/20
1/v=-1/20
v=-20cm
m=-v/u=h_/h
-20/20=h_/10
h_=10
Hope u understand
1/f=1/v+1/u F=-10; U=-20
1/v=1/f-1/u
1/v=-1/10+1/20
1/v=-2+1/20
1/v=-1/20
v=-20cm
m=-v/u=h_/h
-20/20=h_/10
h_=10
Hope u understand
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