Lets say i have a convex lens with a focal length of 10 cm. I place a 4 cm long pointed needle ( ie the object )vertically say 25 cm to the left of the lens; call this x = - 25 cm . The base of the needle is on the principal axis (ie x axis ) and the tip of the needle is 4 cm above the principal axis .
From the thin lens equation the real image is located 16.7 cm to the right of the lens ( ie x = + 16.7 cm ) and it is inverted. The height of the image is reduced to 2.67 cm.
If I put a screen at the image location (ie x = 16.7 cm ) I can easily see the focussed inverted image on the screen . I would be looking at the screen from the left side of the screen .
My questions are as follow:
Q1: If I remove the screen and view the image with my eye positioned at +16.7 cm and looking left ( ie back toward the lens) will I see a clear image ? My assumption is no but I am not sure why not except maybe my eye is too close to the real image .
Q2: If instead of positioning my eye at 16.7 cm I back away and view the image from x at least 16.7 + 25 cm = ~ 42 cm. This way my eye is at least 25 cm from the image which I am assuming is now the “new “ object. I select 25 cm because this is the near point for the normal eye . Once again my eye is looking to the left toward the lens. Could I now see the real image clearly ? I believe I should see the image because there are rays diverging from where the real image is focussed. To me this is the same as my observing any object with my eye with light rays reflecting from the object in all directions. My eye would focus the rays and I should see the image on my retina . Similarly could I a take a picture of the image with a camera with the lens say the same 42 cm away ? Seems plausible to me.