Download App

Physics 3 Marks Question

Ray Optics and Optical Instruments · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 13 questions.

Questions

3 Marks Question

🎯

Test yourself on this topic

13 questions · timed · auto-graded

Question 13 Marks
Light incident normally on a plane mirror attached to a galvanometer coil retrace backwards (as shown in Fig.). A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away? Image
Answer
Here we know that the reflected rays are deflected at an angle twice to the angle of rotation of the mirror. From the figure we see that when the mirror is turned from M to M'position by an angle theta = 3.5 deg The reflected ray OB is bent as:
Image
$\angle 2 \theta=2 \times 3.5^{\circ}=7^{\circ}=\angle AOB$.
In right triangle $AOB , \tan 2 \theta=\frac{ AB }{ AO }$
or $\tan 7^{\circ}=\frac{d}{1.5}$
$\begin{aligned} d & =1.5 \times \tan 7^{\circ} \\ & =1.5 \times 0.1228 \\ & =0.1842 m \\ & =18.42 cm .\end{aligned}$

View full question & answer→
Question 23 Marks
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would be the microscope have to be moved to focus on the needle again?
Answer
Position I. When tank filled with water:
Real depth 12.5 cm
Virtual depth = 9.4 cm
Refractive index of water = n = ?
$\therefore$ Refractive index of water $=\frac{\text { Real depth }}{\text { Virtual depth }}$
$=\frac{12.5 cm}{9.4 cm}=1.33$
Position II. When tank is filled by liquid :
Refractive index of liquid = n = 1.63
Real depth 12.5 cm
Virtual depth = ?
$\therefore$ Refractive index of liquid $=1.63=\frac{12.5}{\text { Virtual depth }}$
$\therefore \quad$ Virtual depth $=\frac{12.5}{1.63}=7.67 cm$
Therefore, the distance through which the microscope has to be moved up
= 9.4-7.67 ⇒ 1.73 cm
$\simeq$1.70 cm
View full question & answer→
Question 33 Marks
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Answer
Given that:
Focal length of the objective lens $=f_{ o }$
$f_o=140 cm$
Focal length of eye lens $=f_e$
Therefore, fe = 5.0 cm
(a) The value of magnifying power is in the normal state of the telescope i. e. the final image is formed at infinity.
$m=\frac{f_o}{|f|}=\frac{140}{5}=28$
(b) When at the minimum (b) When at the minimum distance of distinct vision, then the value of magnifying power will be
$m=\frac{f_o}{\left|f_e\right|}=\left(1+\frac{f_e}{ D }\right)$.
$=\frac{140}{5}\left(1+\frac{5}{25}\right)$
$=28 \times \frac{6}{5}=33.6$
View full question & answer→
Question 43 Marks
Answer the following questions: When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer
The image of the objective of the eyepiece is called exit aperture. All the rays coming from the object pass through the exit aperture after refraction from the objective. Therefore, this is an ideal condition for seeing with our eyes. If we keep our eyes too close to the eyepiece, the eyepiece will not be able to receive much light and the area of vision will also reduce. If we place our eyes on the exit aperture and the area of the pupil of our eye is greater than or equal to the area of the exit aperture, then our eyes will capture all the rays refracted from the objective. The exact location of the aperture generally depends on the distance between the objective and the eyepiece. When we look through a microscope by placing our eye on one end of it, the ideal distance between the eye and the eyepiece is inherent in the design of the instrument.
View full question & answer→
Question 53 Marks
A small pin fixed on a table top is viewed from above a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Answer
Given that:
Thickness of glass slab = t = 15 cm
Refractive index of glass = n = 1.5
Normal change in position of pin = d
Given by $d=t\left(1-\frac{1}{n}\right)$
Here, t = real depth of pin = Thickness of glass slab Let virtual depth of pin = y
Perpendicular change (Value of displacement)
$=t-y=?=d$
$n=\frac{t}{y}$, or $y=\frac{t}{n}$
$\therefore d=$ Perpendicular displacement $=t-\frac{t}{n}$
$=t\left(1-\frac{1}{n}\right)=15\left(1-\frac{1}{1.5}\right)=15\left(1-\frac{2}{3}\right)$
$=15 \times \frac{1}{3}=5 cm$
i.e. the pin will appear to be raised by 5 cm.
No, for small angles of incidence the answer is not dependent on the position of the glass block.
View full question & answer→
Question 63 Marks
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer
Given that :
Focal length of objective lens of telescope $=f_o$
$\therefore \quad f_o=144 cm$
Focal length of eyepiece of telescope $=f_e$
$\therefore \quad f_e=6.0 cm$
Magnification power of telescope $=m=$ ?
Distance between eyepiece and objective lens $= L =$ ?
We know that
Magnification power of telescope $=m=\frac{-f_o}{f_o}$
$m=-\frac{144}{6}=-24$
The separation distance between objective and eyepiece
$L =f_o+f_e$
$=144+6=150 cm$
View full question & answer→
Question 73 Marks
(a) When a telescope is turned upside down and looked at the objective, it appears very small, why?
(b) Why does this not happen in a microscope?
Answer
(a) In a telescope, the focal length $\left(f_o\right)$ of the objective lens is much more than the focal length $\left(f_e\right)$ of the eyepiece and its magnification power is $f_o / f_e$. On looking back, the magnification power will become $f_e / f_o$, because $f_o \ll f_e$ hence now the object will appear very small.
(b) The formula for the magnifying power of a compound microscope is $\frac{v_o}{u_o} \times \frac{ D }{f_e}$, because the value of $v _0$ is only slightly more than the focal length $f_o$ of the objective of the microscope, hence the magnification can be considered as $\frac{v_o}{f_o} \times \frac{ D }{f_e}$; because $f_o$ and $f_e$ both have low values. Therefore, even after turning the microscope, the magnification power remains almost unchanged due to there being no significant difference in the value of $v_0$.
View full question & answer→
Question 83 Marks
What is meant by pure and impure spectrum?
Answer
Pure spectrum : When rays of white light pass through a prism, each ray forms a spectrum on the screen. If each colour is seen separately in the spectrum, then this type of spectrum is called 'pure spectrum'. The following are the conditions for obtaining a pure spectrum :
(i) The line aperture should be narrow and located at the focus of the converging lens.
(ii) The prism should be in a position of minimum deviation.
(iii) Rays of the same colour emanating from the prism should be focused at one place by the converging lens.
(iv) The incident rays should be parallel.
Impure spectrum : When rays of white light pass through a prism then each ray produces a spectrum on the screen. Due to overlapping of different colors, these colours are not visible separately on the screen. This type of spectrum is called 'impure spectrum'.
View full question & answer→
Question 93 Marks
A mobile phone is placed on the principal axis of a concave mirror. Show the formation of the image with a suitable ray diagram. Explain why the magnification is not uniform. Does the distortion of the image depend on the position of the phone relative to the mirror?
Answer

Image
The figure shows a light-ray diagram of the formation of the image of a phone. The image of the part situated in the plane perpendicular to the principal axis will be in the convex hull. It will be of the same size, i.e. $B ^{\prime} C = BC$.
View full question & answer→
Question 103 Marks
Prove that $n_{12}=\frac{1}{\sin i_C}$ where $I_c$ is the critical angle.
Answer
The angle of incidence whose corresponding angle of refraction is $90^{\circ}$. It is called the critical angle $i_c$ for a given pair of mediums. From Snell's law $n_{21}=\frac{\sin i}{\sin r}$ we see that if the relative refractive index is less than one, then the maximum value of $\sin r$ is one. Therefore, there is an upper limit to the value of $\sin i$ up to which this rule can be applied.
This is $i=i_{ C }$, thus $\sin i_{ C }=n_{21}$
Snell's law of refraction cannot be applied to values of $i_c$ greater than $i$. Hence no refraction is possible. The refractive index of denser medium 2 will be relative to rarer medium 1.
$
n_{12}=\frac{1}{\sin i_C}
$
View full question & answer→
Question 113 Marks
Write the definition of wavefront. Using Huygens' principle, draw the shape of a plane wave incident on a convex lens and the refracted wave body.
###
(a) When a wave propagates from a rarer medium to a denser medium, then which characteristic of that wave does not change and why?
(b) The refractive indices of two mediums are $\mu_1$ and $\mu_2$, what will be the ratio of velocities in the wave in them?
Answer
When waves emerge from a wave source located in a medium, then the particles of the medium located around it start vibrating. The surface in the medium in which all the particles located in the same phase of vibration are called wave body.

Image
or
(a) The frequency remains constant because the number of waves entering the two mediums per second is equal.
(b) The value of velocity in different mediums can be given as follows:$
\begin{aligned}
c_1 & =\frac{c_0}{\mu_1} \text { and } c_2=\frac{c_0}{\mu_2} \\
\therefore \quad \frac{c_1}{c_2} & =\frac{\mu_1}{\mu_2}
\end{aligned}
$
In this way the speed of light is inversely proportional to the refractive index of the medium.
View full question & answer→
Question 123 Marks
(a) Stars located very far away, which cannot be seen with the eye, can be seen through a telescope. Why?
(b) The magnifying power of two telescopes is the same but the apertures of their objective lenses are different. What will be the change in the final images formed by them?
Answer
(a) The aperture of the lens of the eye is very small and the light coming from a star located very far away is very less and is unable to stimulate the retina of our eye. But the aperture of the telescope is much larger than that of the eye. Due to this, it can receive proper light from the distant star and forms a bright image of the star, which can be seen.
(b) The brightness of the images will varying; because it depends on the diameter of the aperture. The image formed by a telescope with an objective lens of a larger aperture will be brighter. Also, the resolving power $\left(\frac{ D }{1.22 \lambda}\right)$ of this telescope will also be higher than the smaller aperture of a telescope with an objective lens.
View full question & answer→
Generate Paper FreeAll Ray Optics and Optical Instruments topics
3 Marks Question - Physics STD 12 Science Questions - Vidyadip