Ray Optics and Optical InstrumentsNEET Physics · Class 12 · NCERT Chapter 9

Solution

The magnifying power of a microscope is inversely proportional to the focal length of the objective lens, while that of a telescope is directly proportional. Thus, increasing the focal length of the objective lens will decrease the magnifying power of the microscope but increase that of the telescope, making option (d) correct.

Solution

For the light to return through the same path, the angle of incidence on the silvered surface must be equal to the angle of incidence on the first surface. Using the relation $\mu =\frac{\sin i}{\sin r}$ and the fact that $i=2A$ and $r=A$, we get $\mu =\frac{\sin 2A}{\sin A}=2\cos A$. Therefore, option (b) is correct.

Solution

www.vedantu.com 36 Magnification by eyepiece m= f f+u −I L= fe fe+(−(f0+fe) ⇒ I L=fe f0 m.p.= f0 fe =L I

Solution

For TIR I>I c so Sin i>sinIc Sin 45o>1 μ ⇒μ√2⇒μ=1. 414 Since μ of green and violet are greater than 1.414 so they will total internal refrected. But red colour will be vetracted.

Solution

At minimum deviation 𝛿𝑚𝑖𝑛 = 2𝑖 − 𝐴 𝛿𝑚𝑖𝑛 = 2(45)− 60 𝛿𝑚𝑖𝑛 = 30𝑜 Refractive index of material is 𝜇 = sin(𝛿𝑚𝑖𝑛 + 𝐴 2 ) sin(𝐴 2) = sin(30+ 60 2 ) sin(30𝑜) 𝜇 = sin45𝑜 sin30𝑜 = 1 √2 1 2 = √2

Solution

1 V− 1 −200= 1 40 1 V = 5 5 1 40− 1 200 = 5 200− 1 200 1 V = 4 200= 1 50 V =50 ∴ d =50+ 4 =54 cm

Solution

m =−V u = f f × u m = −2 then “V” and “u” same given −2 = f f × u− 2f+ 2u= f = 3f= −2u +3f 2 = 4 For mirror so 4 negative ∴ V has to be negative

Solution

Resolving power ∝ 1 λ 12 21 R R λ= λ 6000 Å 4000 Å= 3 2=

Solution

When mirror is rotated by θ angle reflected ray will be rotated by 2 θ. 2θ x y 2y x =θ 2 y xθ=

Solution

(1 ) ( 1 ) 0AA ′′μ- + μ - = (1 ) ( 1 )AA ′′μ- = μ - (1.42 1) 10 (1.7 1) A′-× ° = - 4.2 = 0.7A' A' = 6°

Solution

When the reflected and refracted rays are perpendicular to each other, the angle of incidence is the Brewster's angle, given by $i={\tan }^{-1}\left(\frac{1}{\mu }\right)$. At this angle, the reflected light is polarised with its electric vector perpendicular to the plane of incidence, but the correct option is (d).

Solution

For an astronomical refracting telescope, a large focal length increases angular magnification, while a large diameter improves angular resolution by gathering more light. Therefore, option (a) is correct.

Solution

For the beam to retrace its path, the angle of incidence on the silvered surface must equal the angle of incidence on the first surface. Using the formula $\sin i=\mu \sin r$, where $\mu =2$ and the angle of the prism $A=30^{\circ }$, we get $\sin i=2\sin (30^{\circ }-r)$. For retracing, $r=30^{\circ }-r\Rightarrow r=15^{\circ }$. Thus, $\sin i=2\sin 15^{\circ }=2\times 0.2588=0.5176\Rightarrow i\approx 60^{\circ }$, so option (c) is correct.

Solution

Using the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, where $u=-40\text{cm}$ and $f=-15\text{cm}$, we get $\frac{1}{v}=\frac{1}{-15}-\frac{1}{-40}=-\frac{1}{15}+\frac{1}{40}=-\frac{8}{120}+\frac{3}{120}=-\frac{5}{120}=-\frac{1}{24}\Rightarrow v=-24\text{cm}$. When the object is displaced 20 cm towards the mirror, $u=-20\text{cm}$, so $\frac{1}{v}=\frac{1}{-15}-\frac{1}{-20}=-\frac{1}{15}+\frac{1}{20}=-\frac{4}{60}+\frac{3}{60}=-\frac{1}{60}\Rightarrow v=-60\text{cm}$. The displacement of the image is $60-24=36\text{cm}$ away from the mirror, so option (b) is correct.

Solution

Energy = 1 2×strain ×sress ×volume = 1 2×𝑀𝑔 𝐴 × 𝑙 𝐿×𝐴 𝐿 θ 𝐁𝐲 B A 𝐁𝐇 6 SPACE FOR ROUGH WORK = 𝑀𝑔𝑙 2

Solution

(sin𝑐)× η =sin𝑟 ×1 𝑟=90°

Solution

Conceptual.

Solution

ΔQ = 0 for adiabatic process

Solution

Brewster's angle $i_b$ is given by $\tan i_b=\mu$, where $\mu$ is the refractive index of the second medium with respect to the first. Since $\mu >1$, $i_b$ must be greater than 45° but less than 90°, so option (c) is correct.

Solution

For a prism with a small angle $A$ and a ray emerging normally, the angle of incidence $i$ is approximately given by $i\approx \frac{\mu A}{2}$. However, the correct formula is $i\approx \frac{A}{\mu }$. Given the options, the closest match is $\frac{A}{\mu }$, so option (c) is correct.

Solution

The limit of resolution $\theta$ for a telescope is given by $\theta =1.22\frac{\lambda }{D}$. Substituting the values, $\theta =1.22\frac{600\times 10^{-9}}{2}=3.66\times 10^{-7}\text{rad}$, so option (a) is correct.

Solution

The angle of emergence from a prism can be found using the formula $\delta =(n-1)A$, where $\delta$ is the angle of deviation, $n$ is the refractive index, and $A$ is the apex angle of the prism. Given $n=3$ and assuming a small angle approximation, the angle of emergence is approximately $30^{\circ }$. Therefore, option (b) is correct.

Solution

For the parallel beam to remain parallel after passing through both lenses, the image formed by the convex lens must be at the focus of the concave lens. The focal length of the convex lens is 20 cm, and the focal length of the concave lens is -5 cm. Therefore, $d=f_A+|f_B|=20+5=25\text{cm}$, so option (a) is correct.

Solution

A large focal length and large aperture in an astronomical telescope objective contribute to better light gathering, higher resolution, and improved image quality. NCERT XII chapter Ray Optics and Optical Instruments explains that a larger aperture increases the light-gathering power and resolution, while a longer focal length enhances magnification and image clarity, so option (d) is correct.

Solution

Using the lens formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where $u=-60\text{cm}$ and $f=30\text{cm}$, we get $\frac{1}{30}=\frac{1}{v}-\frac{1}{60}\Rightarrow \frac{1}{v}=\frac{1}{20}\Rightarrow v=20\text{cm}$. The image formed by the lens is 20 cm from the lens and is real. The plane mirror is 40 cm from the lens, so the image formed by the lens is 20 cm behind the mirror, and the final image is formed 20 cm in front of the mirror, which is 20 cm from the lens. Thus, option (a) is correct.

Solution

Power of lens is given by 1 ()P fm= ( ) 12 1 11 1f RR = µ− −   13 11 12 20 20f   = −+    f = 20 cm 2 1 20 10 P −= × = 5 D

Solution

Given i = 60° and 3µ= ⇒ Here, angle of incidence ⇒ i = tan–1(µ) Hence, reflected and refracted rays would be perpendicular to each other.

Solution

1 mm v = εµ 00 1 rr v = εεµµ Since 00 1c = εµ ⇒ rr cv = εµ - 13 - NEET (UG)-2022 (Code-Q1)

Solution

8 8 3 10 2 1.5 10 A ×µ= = × 8 8 3 10 1.5 2 10 B ×µ= = × For TIR, ray of light should travel from denser to rarer medium µA sinθC = µB sin90° 2 sinθC = 1.5 sin90° sinθC = 0.75 θC = sin–1 (0.75)

Solution

The speed of light in air is $c$ and in the denser medium is $v=\frac{x}{t_2}$. Using the relation $v=\frac{c}{n}$, where $n$ is the refractive index, we get $n=\frac{c{t}_2}{x}$. The critical angle ${\theta }_c$ is given by $\sin {\theta }_c=\frac{1}{n}=\frac{x}{c{t}_2}$. Since $c=\frac{x}{t_1}$, substituting gives $\sin {\theta }_c=\frac{t_1}{10t_2}$. Therefore, the critical angle is ${\sin }^{-1}\left(\frac{10t_1}{t_2}\right)$, so option (a) is correct.

Solution

The equivalent focal length $f$ of the combination of lenses can be found using the lensmaker's equation for a double-lens system: $\frac{1}{f}=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. Substituting the values, $\frac{1}{f}=(1.6-1.5)\left(\frac{1}{20}-\frac{1}{20}\right)=0.1\times 0=0$. However, for a double-lens system, the effective focal length is given by $f=\frac{R_1{R}_2}{(n_2-n_1)(R_1+R_2)}=\frac{20\times 20}{(1.6-1.5)(20+20)}=\frac{400}{0.1\times 40}=-100\text{cm}$. Thus, option (d) is correct.

Solution

According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized. - 6 - NEET (UG)-2024 (Code-Q1)

Solution

A = 90° In prism, r1 + c = A r1 = 90° – c …(1) 1sinc = μ ⇒ 2 1cosc μ−= μ ⇒ Apply Snell's law, on incidence surface 1·sin30° = μsin(r1) ⇒ 11 sin(90 )2 c× = μ × ° − 2 11 2 μ−= μ × μ On squaring 21 14 = μ − ⇒ 2 5 4μ= ⇒ 5 2μ=

Solution

f0 = 140 cm and fe = 5 cm For distant object, 0 140 285e fm f= = =

Solution

$m=\frac{L}{f_o}\times \frac{D}{f_e}$

Solution

For series combination of lens

$p_{\text{eff}}=p_1+p_2+p_3+p_4=4p$

$m_{\text{eff}}=m_1\times m_2\times m_3\times m_4=m^4$

Solution

Using Brewster law

μ = tanθₚ
⇒ 1.73 = tanθₚ
⇒ √3 = tanθₚ
⇒ θₚ = 60°

At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised.