Question

(I) The real image of an object placed between $f$ and $2f$ from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? Explain.

(II) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

Hint:

Key ideas:

Real image exists even without a screen.
Plane mirror → always virtual image.
Convex mirror → real image possible only for virtual objects.

Answer 1

Step 1: Part (I), image position and nature:
The object lies between \( f \) and \( 2f \) of the convex lens.
The lens forms a real, inverted and magnified image beyond \( 2f \).

Step 2: Part (I), role of the screen:
A real image is a point where refracted rays actually meet.
The screen only scatters these rays towards the eye and makes the image visible.
It does not create the image.

Step 3: Part (I), conclusion:
If the screen is removed, the rays still meet at the same point and then spread out.
So the image is still there in space.
An eye placed behind the image position, looking towards the lens, receives the diverging rays and sees the image.

Step 4: Part (II), plane mirror:
A real object always gives a virtual image in a plane mirror.
The reflected rays diverge, so they cannot meet in front of the mirror.
Now let a converging beam from a lens fall on the mirror. It would have met at a point behind the mirror.
This point is a virtual object.
The plane mirror gives \( v = -u \) in magnitude, so the reflected rays meet at the mirror image of that point, in front of the mirror.
This is a real image.

Step 5: Part (II), convex mirror:
A convex mirror also gives a virtual, erect and diminished image of a real object.
For a virtual object at distance \( u \) behind the mirror, the mirror formula gives \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \) with \( f \) positive.
If the virtual object lies between the pole and the focus, that is \( u < f \), then \( v \) is negative.
Example: \( f = 20 \) cm and \( u = 10 \) cm give \( v = \frac{fu}{u-f} = -20 \) cm.
The image is real and lies 20 cm in front of the mirror.

Final Answer:
(I) Yes, the real image is still there after the screen is removed.
(II) Yes. A plane mirror or a convex mirror forms a real image when a converging beam falls on it, that is, when the object is virtual.