Ray Optics MCQ Questions & Answers in Optics and Wave | Physics
Learn Ray Optics MCQ questions & answers in Optics and Wave are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
221.
A telescope has an objective lens of focal length $$150\,cm$$ and an eyepiece of focal length $$5\,cm.$$ If a $$50\,m$$ tall tower at a distance of $$1\,km$$ is observed through this telescope in normal setting, the angle formed by the image of the tower is $$\theta ,$$ then $$\theta $$ is close to:
222.
In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $$u$$ and the image distance $$v,$$ from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the $$x$$ - axis meets the experimental curve at $$P.$$ The coordinates of $$P$$ will be:
A
$$\left( {\frac{f}{2},\frac{f}{2}} \right)$$
B
$$\left( {f,f} \right)$$
C
$$\left( {4\,f, 4\,f} \right)$$
D
$$\left( {2\,f, 2\,f} \right)$$
Answer :
$$\left( {2\,f, 2\,f} \right)$$
Here $$u = - 2\,f, v = 2\,f$$
As $$\left| u \right|$$ increases, $$v$$ decreases for $$\left| u \right|$$ $$> f.$$ The graph between $$\left| v \right|$$ and $$\left| u \right|$$ is shown in the figure. A straight line passing through the origin and making an angle of 45° with the $$x$$ - axis meets the experimental curve at $$P(2\,f, 2\,f).$$
223.
A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is $$\frac{2}{3}.$$ Their equivalent focal length is $$30\,cm.$$ What are their individual focal lengths?
224.
The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is $$40\,cm.$$ The area of the image is 9 times that of the square. The focal length of the lens is :
A
$$36\,cm$$
B
$$27\,cm$$
C
$$60\,cm$$
D
$$30\,cm$$
Answer :
$$30\,cm$$
If side of object square $$ = \ell $$
and side of image square $$ = \ell '$$
From question, $$\frac{{\ell '}}{\ell } = 9\,\,{\text{or}}\,\,\frac{{\ell '}}{\ell } = 3$$
i.e, magnification $$m = 3$$
$$u = - 40\,cm$$
$$v = 3 \times 40 = 120\,cm\,f = ?$$
From formula, $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{
& \frac{1}{{120}} - \frac{1}{{ - 40}} = \frac{1}{f}\,\,{\text{or,}}\,\,\frac{1}{f} = \frac{1}{{120}} + \frac{1}{{40}} = \frac{{1 + 3}}{{120}} \cr
& \therefore f = 30\,cm \cr} $$
225.
Which of the following is used in optical fibres?
A
total internal reflection
B
scattering
C
diffraction
D
refraction
Answer :
total internal reflection
In an optical fibre, light is sent through the fibre without any loss by the phenomenon of total internal reflection as shown in the figure.
226.
A transparent cube contains a small air bubble. Its apparent distance is $$2\,cm$$ when seen through one face and $$5\,cm$$ when seen through other face. If the refractive index of the material of the cube is $$1.5,$$ the real length of the edge of cube must be
A
$$7\,cm$$
B
$$7.5\,cm$$
C
$$10.5\,cm$$
D
$$\frac{{14}}{3}cm$$
Answer :
$$10.5\,cm$$
As light travels from denser to rarer medium, so refractive index $$\left( \mu \right) = \frac{{{\text{real depth}}}}{{{\text{apparent depth}}}}$$
Refractive index $$\left( \mu \right) = 1.5$$
Net apparent depth $$ = 2 + 5 = 7\,cm$$
And real depth $$ = {\text{apparent depth}} \times \mu $$
$$\therefore $$ Real depth $$ = 1.5 \times 7 = 10.5\,cm$$