# 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

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.

#### 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.

Is there an error in this question or solution?
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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