A rod of length 2 text{ cm} makes an angle of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rod makes an angle of $\frac{2 \pi}{3} 	ext{ rad}$ $(120^{\circ})$ with the principal axis. The angle with the negative x-axis is $60^{\circ}

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The rod makes an angle of $\frac{2 \pi}{3} 	ext{ rad}$ $(120^{\circ})$ with the principal axis. The angle with the negative x-axis is $60^{\circ}$. The horizontal component of the rod is $L_x = 2 \cos(60^{\circ}) = 1 	ext{ cm}$ and the vertical component is $L_y = 2 \sin(60^{\circ}) = \sqrt{3} 	ext{ cm}$.The distance of the object from the lens is $u = -\frac{40}{3} 	ext{ cm}$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{10} - \frac{3}{40} = \frac{4-3}{40} = \frac{1}{40}$,so $v = 40 	ext{ cm}$.The lateral magnification is $m = \frac{v}{u} = \frac{40}{-40/3} = -3$.The height of the image is $h_i = |m| \cdot L_y = 3 \cdot \sqrt{3} = 3\sqrt{3} 	ext{ cm}$.The longitudinal magnification is $m_L = -\frac{v^2}{u^2} = -(\frac{40}{-40/3})^2 = -9$.The length of the image along the principal axis is $L'_x = |m_L| \cdot L_x = 9 \cdot 1 = 9 	ext{ cm}$.The angle $\alpha$ made by the image with the principal axis is given by $	an \alpha = \frac{h_i}{L'_x} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Thus,$\alpha = 30^{\circ} = \frac{\pi}{6} 	ext{ rad}$.Therefore,$n = 6$.

Similar Questions

The focal length of a convex lens of refractive index $1.5$ is $2 \ cm$. What will be the focal length of the lens when it is immersed in a liquid of refractive index $1.25$?

$A$ convex lens of focal length $0.15 \,m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \,m$. The refractive index of the liquid is

The focal length of a thin biconvex lens is $20 \,cm$. When an object is moved from a distance of $25 \,cm$ in front of it to $50 \,cm$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module