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# The focal length of a lens is 0.12 m and the Refractive Index (R.I.) is 1.5. The focal length of the same lens for blue colour is 0.1 m. The R.I. of the lens for blue colour is(A) 1.5(B) 1.25(C) 1.45(D) 1.6

Last updated date: 09th Sep 2024
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Hint: We know that in physics, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium. Refraction of light is the most commonly observed phenomenon, but other waves such as sound waves and water waves also experience refraction. Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. The wavelength decreases as the light enters the medium and the light wave changes direction.

We know that for a thin lens in air, the focal length is the distance from the center of the lens to the principal foci (or focal points) of the lens. For a converging lens (for example a convex lens), the focal length is positive, and is the distance at which a beam of collimated light will be focused to a single spot. Focal length can also change the perspective and scale of our images. A lens with a shorter focal length expands perspective, giving the appearance of more space between the elements in your photo. Meanwhile, telephoto lenses tend to stack elements in the frame together to compress perspective.
We know that lenses with short focal lengths will have a wide angle of view, while longer focal length lenses will have stronger magnification, creating what is called a telephoto lens. As the focal length value decreases, the lens becomes a wide-angle lens, and as the value increases it becomes a telephoto lens.
$\mathrm{f}=\mathrm{m}=12 \mathrm{cm}$
$\mu=1.5$
$\mathrm{f}_{\mathrm{b}}=10 \mathrm{cm}$
So, the formula we get is:
$\dfrac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\dfrac{\left(\mu_{2}-1\right)}{\left(\mu_{1}-1\right)}$
Now we have to put the values to get the value as:
$\dfrac{12}{10}=\dfrac{\left(\mu_{\mathrm{b}}-1\right)}{(0.5)}$
$\dfrac{12 \times 5}{100}=\mu_{\mathrm{b}}-1$
Hence, after the evaluation we get that:
$\mu_{\mathrm{b}}=1.6$

Hence the correct answer is option B.

Note: We know that refractive index, also called index of refraction, measures the bending of a ray of light when passing from one medium into another. Refractive index has a large number of applications. It is mostly applied for identifying a particular substance, confirming its purity, or measuring its concentration. Generally, it is used to measure the concentration of a solute in an aqueous solution. The higher the refractive index the slower the light travels, which causes a correspondingly increased change in the direction of the light within the material. What this means for lenses is that a higher refractive index material can bend the light more and allow the profile of the lens to be lower.