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Ray Optics and Optical Instruments

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2 months ago

Ray Optics and Optical Instruments Class 12th

Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt[]{2 n} .$ The incident angle is twice that of refracting angle. Then, the angle of incidence will be:

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Vishal Baghel

Contributor-Level 10

Given I = 2r

$μ_{1} = 1 (A i r)$

$μ_{2} = \sqrt[]{2 n}$

by, Snell's law

$μ s i n i = μ_{2} s i n r$

$i = 2 c o s^{- 1} [\sqrt[]{\frac{n}{2}}]$

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2 months ago

Ray Optics and Optical Instruments Class 12th

A thin lens made of glass (refractive index $= 1.5$ ) of focal length $f = 16 c m$ is immersed in a liquid of refractive index 1.42. If its focal length in liquid is $f_{l}$ , then the ratio $f_{l}$ /f is closest to the integer:

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alok kumar singh

Contributor-Level 10

$\frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f_{l}} = (\frac{μ_{g}}{μ_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

(i) $\frac{f_{l}}{f} = \frac{μ_{g} - 1}{\frac{μ_{g}}{μ_{l}} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.41} - 1} = 8.875 \approx 9$

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2 months ago

Ray Optics and Optical Instruments Class 12th

Time taken by light to travel in two different materials A and B refractive indices $μ_{A} a n d μ_{B}$ of same thickness is $t_{1} a n d t_{2}$ respectively. It $t_{2} - t_{1} = 5 * 1 0^{- 10} s$ and the ratio of $μ_{A} t o μ_{B}$ 1 : 2. Then, the thickness of material, in meter is: (Given $υ_{A} a n d υ_{B}$ are velocities of light in A and B materials respectively.)

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Vishal Baghel

Contributor-Level 10

$\frac{μ_{A}}{μ_{B}} = \frac{C / v_{A}}{C / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2}$

Let the thickness is d

$d = \frac{5 * 1 0^{- 10} * v_{A} v_{B}}{v_{A} - v_{B}}$

d = 5 * 10^-10 * v_A

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2 months ago

Ray Optics and Optical Instruments Class 12th

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $λ,$ calculate the change of microscope's resolving power due to oil and choose the correct option

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alok kumar singh

Contributor-Level 10

Resolving power

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New answer posted

2 months ago

Ray Optics and Optical Instruments Class 12th

For an object placed at a distance 2.4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm from the lens. A glass plate of refractive index 1.5 and thickness 1cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

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Payal Gupta

Contributor-Level 10

When glass slab is not put in between lens and screen

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{12} - \frac{1}{- 240} = \frac{1}{f} . . . . . . (1)$

When glass slab is put in between lens and screen

Shift = $t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} c m$

$\Rightarrow \frac{3}{35} - \frac{1}{- x} = \frac{1}{f} . . . . . . . . . (2)$

With the help of equations (1) and (2), we can write

$\frac{3}{35} - \frac{1}{- x} = \frac{1}{12} - \frac{1}{- 240}$

$\Rightarrow | Δ u | = 560 - 240 = 320 c m = 3.2 m$

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2 months ago

Ray Optics and Optical Instruments Class 12th

Light enters form air into a given medium at angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is:

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alok kumar singh

Contributor-Level 10

Using snell's law –

sin I = $? s i n ? r$

->sin45° = $? s i n 30 °$

$? = \frac{s i n 45}{s i n 30} = \frac{1}{\sqrt[]{2}} * 2 = \sqrt[]{2}$

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2 months ago

Ray Optics and Optical Instruments Class 12th

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

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Vishal Baghel

Contributor-Level 10

Angular magnification, $θ = | \frac{f_{0}}{f_{e}} |$ $\frac{_{}}{_{}}$

Length of tuber L = $| f_{0} | + | f_{e} |$ $_{}_{}$

$| f_{0} | + | f_{e} | = 30 c m$

$A s, | \frac{f_{0}}{f_{e}} | = 2$

$| f_{0} | = 2 | f_{e} |$

$\Rightarrow | f_{0} | + \frac{| f_{0} |}{2} = 30 \Rightarrow | f_{0} | = \frac{2}{3} * 30 = 20 c m$

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2 months ago

Ray Optics and Optical Instruments Class 12th

As shown in the figure, after passing through the medium 1. The speed of light $υ_{2}$ in medium 2 will be:

(Given c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

In a medium, speed of light, $V = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8} m / s}{\sqrt[]{1 * 9}} = 1 * 1 0^{8} m / s$

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2 months ago

Ray Optics and Optical Instruments Class 12th

A thin prism of angle 6° and refractive index for yellow light (n_Y) 1.5 is combined with another prism of angle 5° and n_Y = 1.55. The combination produces no dispersion. The net average deviation $(δ)$ produced by the combination is ${(\frac{1}{x})}^{o}$ . The value of x is____________.

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Vishal Baghel

Contributor-Level 10

$δ = A (μ_{y} - 1) - A^{1} (μ_{y}^{1} - 1)$

$= 6 (1.5 - 1) - 5 (1.55 - 1)$

$= \frac{1}{4}$

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New answer posted

3 months ago

Ray Optics and Optical Instruments Class 12th

A convex lens of focal length 20cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ______cm.

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Payal Gupta

Contributor-Level 10

For refraction of light at lens

$\frac{1}{v} - \frac{1}{- 60} = \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} \Rightarrow \frac{1}{v} = \frac{3 - 1}{60} = \frac{1}{30}$

v = 30 cm

If image has to be formed at object itself then light ray should retrace its path. Hence after refraction at lens, it must strikes normally to the mirror

RM = 20 cm Radius of curvature of mirror

Fm = 10 cm focal length of mirror

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