Using the new Cartesian sign convention, we get a positive focal length for the convex lens and a negative focal length for the concave lens.

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Problem 2: A convex lens forms a real and inverted image of an object 40cm from the lens. Where will be the object placed in front of the convex lens, if the image is of the same size as the object?

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We know the properties of convex lens that it is virtual and upright. Because a convex lens may create both virtual and actual pictures, the magnification produced by a convex lens can be either positive or negative. Magnification is beneficial for virtual images but detrimental for real images. i.e. Positive (+ve) for virtual image and Negative (-ve) for the real image.

Magnificationformula for convexlens

Problem 3: A concave lens of the focal length of 20cm forms an image of a needle 15 cm from the lens. How far is the needle placed from the lens?

Case 3: If the magnitude of magnification is one, then the image is the same size as an object. |m| = 1, the image is same size as object.

Light is a kind of energy that can be seen with the naked eye. We observe objects and understand the world around us mostly via the use of light. Light travels in a straight path at an extremely fast speed of around 3 × 108 ms. A small light source produces a strong shadow on an opaque object. This means that the light travels in a straight line and the route is referred to as a ray of light, and a grouping of rays is referred to as a beam of light.

When the item is in front of the concave lens, the image is in front of the same object on the same side. The concave lens always produces a virtual, erect, and reduced image. Because concave lenses always generate virtual images, the magnification achieved by them is always positive and it always produces an image that is smaller than the object.

Problem 5: An optical system uses two thin convex lenses in contact having an effective focal length of 30/4 cm. If one of the lenses has a focal length of 30cm, find the focal length of the other.

Problem 1: The radii of curvatures of a convex lens are 40cm and 50cm, calculate the focal length if the refractive index of its material is 2.1.

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Case 1: If the magnitude of magnification is less than one, it means the image is smaller than the object. |m|<1, the image is diminished.

This is the lens maker`s equation. The lens of any desired focal length can be produced by choosing proper values of R1, R2, μ1, μ2.

Magnificationformula Biology

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Magnification of lens equationexample

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Magnificationformula forlensin termsoffocal length

Problem 4: a concave lens made of glass has a focal length of 20cm in air. Find its focal length when immersed in water. Given that the refracting index of the glass lens is 1.5 and that of in water is 4.

A light ray indicates the direction of light propagation. When light strikes a surface between two transparent mediums, it reflects and refracts, causing light rays to bend. Light rays bend around the edge of obstruction as well, although the bending is relatively minimal due to the very short wavelength of light radiation. This is known as light diffraction.

Below tabular representation indicates the magnification and nature of the image for different cases for different lenses as:

The lens equation is used by lens makers employ to create lenses with desired focal lengths. Lenses with varying focal lengths are employed in a variety of optical devices. The focal length of a lens is determined by the radii of curvature of two surfaces and the refractive index of the lens material.

Magnification of lens equationphysics

Consider a thin glass lens with a refractive index of μ2 and curvature centres C1 and C2 with curvature radii R1 and R2. Let μ1 represent the refractive index of the surrounding medium. When a point object ‘O’ is maintained on-axis at a distance u from the lens, the ray OP passes through the optical centre without deviation. If the other ray OA had not been refracted along with AB by the first surface, it would have arrived at the point I’. If the second surface didn`t exist. But, due to the second surface, the ray undergoes another refraction at point B and reaches point I.

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Magnification ofconvexlensis positive or negative

Magnificationformula for mirror

For the second refraction, I’ acts as a virtual object in lens medium of refractive index μ2, to produce a final image at I in surrounding medium of refractive index μ1. By using the relation among u. v. μ1, and R for refraction at a single curved (convex) surface.

When a thin lens is submerged in water, its relative refractive index decreases, and therefore its focal length increases (and the lens’s power decreases).

Magnification of lens equationderivation

For the first refraction, the object distance is u, and the image distance is v’. By using the relation among u. v. μ1, and R for refraction at a single curved (convex) surface.

by VD Tang · 2023 — Function · Focal length is the distance between the lens and the focal point, where the light rays converge or diverge. · The concept of ...

The focal length of a lens is determined by the refractive index of the lens’s material in relation to its surroundings, as determined by the lens maker’s formula.

It is defined as the distance between the optical center and the second primary focus, so that the focal length of a convex lens is positive and that of a concave lens is negative.

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The image distance can be computed using the lens formula and knowledge of the object distance and focal length. The Lens formula describes the relationship between the distance of an image I the distance of an object (o), and the focal length (f) of the lens in optics. The lens formula works for both convex and concave lenses.

Magnification means making objects appear larger than they are. Following are the different cases to determine the magnification for different cases as:

It is equivalent to the image-to-object distance ratio. *** QuickLaTeX cannot compile formula: *** Error message: Error: Nothing to show, formula is empty