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A convex lens which has focal length of 4 cm and refractive index of 1.4 is immersed in a liquid of refractive index 1.6 .the focal length will be?
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A convex lens which has focal length of 4 cm and refractive index of 1...
Fnew/fair = ug-1/(ug /um-1)

Fn /Fa = 1.4-1/(1.4/1.6 -1) = 8.cm
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A convex lens which has focal length of 4 cm and refractive index of 1...
**Focal Length of the Lens in Air**

The focal length of a lens is a measure of its ability to converge or diverge light. In this case, we have a convex lens with a focal length of 4 cm when it is in air.

**Refractive Index of the Lens Material**

The refractive index of a material determines how light propagates through it. It is calculated as the ratio of the speed of light in a vacuum to the speed of light in the material. In this case, the refractive index of the lens material is given as 1.4.

**Refraction at the Lens-Air Interface**

When light passes from one medium to another with a different refractive index, it undergoes refraction. The amount of refraction is determined by Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media.

In this case, the light is passing from air (refractive index = 1) to the lens material (refractive index = 1.4). Let's assume the angle of incidence is θ1 and the angle of refraction is θ2.

**Refraction at the Lens-Liquid Interface**

After passing through the lens material, the light enters a liquid with a refractive index of 1.6. Let's assume the angle of incidence at this interface is θ3 and the angle of refraction is θ4.

**Lens Maker's Formula**

The lens maker's formula relates the focal length of a lens to the refractive index of the lens material and the radii of curvature of its surfaces. However, in this case, we are not given the radii of curvature, so we cannot directly use the lens maker's formula.

**Using the Lens Formula**

The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of a lens. It is given by:

1/f = 1/v - 1/u

where u is negative for a convex lens.

In this case, the object distance is assumed to be large compared to the focal length, so we can consider the lens to be in the thin lens approximation. Therefore, the lens formula can be simplified to:

1/f = 1/v

**Applying Snell's Law and Lens Formula**

1. From Snell's law, we know that n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the refractive indices of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.

2. Similarly, at the lens-liquid interface, n2 * sin(θ3) = n3 * sin(θ4), where n3 is the refractive index of the liquid.

3. Since the lens is thin, we can consider the angle of incidence at the lens-liquid interface (θ3) to be equal to the angle of refraction at the lens-air interface (θ2).

4. Using the lens formula, we can write 1/f = 1/v = (n2 - 1) * (1/r1 - 1/r2), where r1 and r2 are the radii of curvature of the lens surfaces.

5. Since the lens is symmetrical,
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A convex lens which has focal length of 4 cm and refractive index of 1.4 is immersed in a liquid of refractive index 1.6 .the focal length will be?
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