n object is placed 40 cm from a lens of focal length -10 cm. The image is virtual or real? inverted or upright? diminshed or enlarged?
virtual, upright, and diminished.
I know this is the answer. Would someoen be able to explain why
Well, well, well, let's have a look at this optical conundrum! So, we have an object placed 40 cm from a lens with a focal length of -10 cm. First off, let me remind you that negative focal lengths indicate a diverging lens, also known as a "weirdo" lens. Now, onto your query.
Since the object is further away from the lens than its focal length, it falls into the realm of real images. Sorry, virtual image enthusiasts, not this time! As for the orientation, brace yourself, pal, because the image is inverted. That's right, it's playing peek-a-boo in the mirror! Now, for the cherry on top, the image is also diminished. It's taking a step back in size, trying to be all modest.
So, to summarize: the image is real, inverted, and diminished. Phew! Wasn't that a rollercoaster of optical delight?
To determine whether the image formed by a lens is virtual or real, we need to use the lens formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens,
- v is the image distance from the lens, and
- u is the object distance from the lens.
Given that the object distance (u) is 40 cm and the focal length (f) is -10 cm, we can substitute these values into the lens formula:
1/(-10) = 1/v - 1/40
Simplifying this equation will give us the image distance (v). Please give me a moment to calculate it for you.
To determine whether the image is virtual or real, inverted or upright, and diminished or enlarged, we can use the lens formula and the sign conventions used in optics.
The lens formula is:
1/f = 1/u + 1/v
where:
- f is the focal length of the lens
- u is the object distance from the lens (positive for object on the same side as the incident light)
- v is the image distance from the lens (positive for a real image on the opposite side of the lens, negative for a virtual image)
In this case, the object distance (u) is +40 cm (since it is on the same side as the incident light), and the focal length (f) is -10 cm (negative because it's a concave lens).
By substituting these into the lens formula, we get:
1/-10 = 1/40 + 1/v
Simplifying the equation, we have:
-1/10 = 1/40 + 1/v
To find out whether the image is virtual or real, inverted or upright, and diminished or enlarged, we need to find the value of v.
By solving the equation, we get:
v = -20 cm
Now, let's interpret the values of v and answer your questions:
1. The negative value of v (-20 cm) indicates that the image is virtual because it is on the same side as the object (incident light) and cannot be obtained on a screen.
2. Since the image is virtual and on the same side as the object, it is upright.
3. The image distance (-20 cm) is smaller than the object distance (40 cm), indicating that the image is diminished, meaning it is smaller in size than the object.
To summarize:
The image is virtual, upright, and diminished.