Use the Mirror Equation to Show that an Object Placed Between F and 2f of a Concave Mirror Produces a Real Image Beyond 2f. - Physics

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Numerical

Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

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Solution

According to the mirror equation, we have

`1/"u" + 1/"v" = 1/"f"`

where
u = Distance of object from the mirror
v = Distance of image from the mirror 
f  = Focal length of the mirror

From mirror equation, we have

`"v" = ("uf")/("u" - "f")` ...........(i)

Applying new cartesian sign convention, we get

f = ve and u = ve

Given: f < u < 2f

v = ve .....................[From (i)]

Magnification is given by

`"m" = -(-"v")/(-"u")` = ve

Hence, the image formed is real.

From mirror formula, when u = 2f,

`1/(-2"f") + 1/"v" = 1/-"f"`

`1/"v" = 1/(2"f") - 1/"f" = -1/(2"f")`

Therefore, when the object is at 2f, the image is formed at 2f

Now, when u = f

`1/-"f" + 1/"v" = -1/"f"`

`1/"v" = 1/0`

v = ∞

Therefore, when the object is at f, the image is formed at infinity.

This shows that when f < u < 2f, < v < 2f.

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Chapter 9: Ray Optics and Optical Instruments - Exercise [Page 345]

APPEARS IN

NCERT Class 12 Physics (Part 1 and 2)
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.15 (a) | Page 345
NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 15.1 | Page 346

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