MCQ 11 Mark
When light enters the atmosphere it strikes on extremely fine particles, which deflect the rays of light in all possible directions, This is due to -
MCQ 21 Mark
An object is placed in front of a convex mirror. Its image is formed :
- A
at a distance equal to the object distance in front of the mirror.
- B
at twice the distance of the object in front of the mirror.
- C
half the distance of the object in front of the mirror.
- ✓
behind the mirror and it's position varies according to the object distance.
AnswerCorrect option: D. behind the mirror and it's position varies according to the object distance.
(D) behind the mirror and its position varies according to the object distance.
View full question & answer→MCQ 31 Mark
Study the given ray diagrams and select the correct statement from the following :
- A
Device $X$ is a concave mirror and device $Y$ is a convex lens, whose focal lengths are $20 \ cm$ and $25 \ cm $ respectively.
- B
Device $X$ is a convex lens and device $Y$ is a concave mirror, whose focal lengths are $10 \ cm$ and $25 \ cm $ respectively.
- C
Device $X$ is a concave lens and device $Y$ is a convex mirror, whose focal lengths are $20 \ cm$ and $25 \ cm$ respectively.
- ✓
Device $X$ is a convex lens and device $Y$ is a concave mirror, whose focal lengths are $20 \ cm$ and $25 \ cm$ respectively.
AnswerCorrect option: D. Device $X$ is a convex lens and device $Y$ is a concave mirror, whose focal lengths are $20 \ cm$ and $25 \ cm$ respectively.
From the given figures we can conclude that the device $X$ is a convex lens as it converges the rays and the device $Y$ is a concave mirror as it also converges the rays but in the same side of the mirror .Their focal lengths are $20 \ cm$ and $25 \ cm$ respectively Regards.
View full question & answer→MCQ 41 Mark
Study the following ray diagram : In this diagram, the angle of incidence, the angle of emergence and the angle of deviation respectively have been represented by.
- A
$\ce{y, p, z}$
- B
$\ce{x, q, z}$
- ✓
$\ce{p, y, z}$
- D
$\ce{p, z, y}$
AnswerCorrect option: C. $\ce{p, y, z}$
The angle between the incident ray and the normal is known as the angle of incidence, and the angle between the emergent ray and the normal is known as the angle of emergence. The emergent ray is bent at an angle with the direction of the incident ray. This angle is called the angle of deviation.
View full question & answer→MCQ 51 Mark
A student very cautiously traces the path of a ray through a glass slab for different values of the angle of incidence refraction $(\angle\text{i}).$ He then measures the corresponding values of the angle of refraction $(\angle\text{r}) $ and the angle of emergence $(\angle\text{e})$ for every value of the angle of incidence. On analysing these measurements of angles, his conclusion would be.
- A
$\angle\text{i} > \angle\text{r} > \angle\text{e}$
- ✓
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
- C
$\angle\text{i} < \angle\text{r} < \angle\text{e}$
- D
$\angle\text{i}=\angle\text{e} < \angle\text{r}$
AnswerCorrect option: B. $\angle\text{i}=\angle\text{e} > \angle\text{r}$
Since the emergent ray is always parallel to the incident ray, therefore the angle of incidence is always equal to the angle of emergence i.e.,
$\angle\text{i}=\angle\text{e}\ .....\text{(i)}$
Also, when a light ray travels from rarer to denser medium, it bends towards the normal. Hence
$\angle\text{i}>\angle\text{r}\ ..... \text{(ii)}$
From $(i)$ and $(ii),$ we get
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
Hence, the correct answer is $\angle\text{i}=\angle\text{e} > \angle\text{r}.$
View full question & answer→MCQ 61 Mark
A student obtains a blurred image of a distant object on a screen using a convex lens. To obtain a distinct image on the screen he should move the lens.
- A
- B
- C
To a position very far away from the screen.
- ✓
Either towards or away from the screen depending upon the position of the object.
AnswerCorrect option: D. Either towards or away from the screen depending upon the position of the object.
The position of object is not known whether it is at infinity or beyond centre of curvature or at centre of curvature of lens. So depending upon the position of object we have to move the lens away or towards the screen.
View full question & answer→MCQ 71 Mark
In your laboratory you trace the path of light rays through a glass slab for different values of angle of incidence $(\angle\text{i})$ and in each case measure the values of the corresponding angle of refraction $(\angle\text{r})$ and angle of emergence $(\angle\text{e}).$ On the basis of your observations your correct conclusion is :
- ✓
$\angle\text{i }$ is more than $ \angle\text{r},$ but nearly equal to $ \angle\text{e}$
- B
$\angle\text{i }$ is less than $\angle\text{r },$ but nearly equal to $\angle\text{e}$
- C
$\angle\text{i }$ is more than $ \angle\text{e } , $ but nearly equal to $\angle\text{r }$
- D
$\angle\text{i }$ is less than $ \angle\text{e },$ but nearly equal to $\angle\text{r} .$
AnswerCorrect option: A. $\angle\text{i }$ is more than $ \angle\text{r},$ but nearly equal to $ \angle\text{e}$
On entering a glass slab, the incident light gets refracted. According to Snell’s law, we get
$\mu=\frac{\sin\text{i}}{\sin\text{r}}$
For glass $\mu > 1$
$\therefore \sin\text{r} < \sin\text{i}$
or $r < i$
In refraction of light through a glass slab, the emergent ray is parallel to the incident ray. Thus, $\angle\text{i}=\angle\text{e}$. View full question & answer→MCQ 81 Mark
In the following ray diagram the correctly marked angle are :
- A
$\angle\text{i}$ and $\angle\text{e}$
- B
$\angle\text{A }$ and $ \angle\text{D}$
- C
$\angle\text{i }, \angle\text{e }$ and $ \angle\text{D}$
- ✓
$\angle\text{r }, \angle\text{A } $ and $\angle\text{D}.$
AnswerCorrect option: D. $\angle\text{r }, \angle\text{A } $ and $\angle\text{D}.$
The angle between the normal and the incident ray is the angle of incidence. The angle between the normal and the emergent ray is the angle of emergence. The correctly marked angles are shown in the diagram below :
View full question & answer→MCQ 91 Mark
A student obtains a sharp image of the distant window $(W)$ of the school laboratory on the screen $(S)$ using the given concave mirror $(M)$ to determine its focal length. Which of the following distances should he measure to get the focal length of the mirror?
- A
$MW.$
- ✓
$MS.$
- C
$SW.$
- D
$MW - MS.$
AnswerTo get the focal length of the mirror, the student should measure the distance between mirror and screen because when the object is at infinity, the image is formed at focus i.e. between screen and mirror. So, the distance between screen and mirror should be measured.
View full question & answer→MCQ 101 Mark
A student traces the path of a ray of light through a rectangular glass slab for the different values of angle of incidence. He observes all possible precautions at each step of the experiment. At the end of the experiment, on analysing the measurements, which of the following conclusions is he likely to draw?
- A
$\angle\text{i} = \angle\text{ e} < \angle\text{r}$
- B
$\angle\text{i} < \angle\text{ e} < \angle\text{r}$
- C
$\angle\text{i} > \angle\text{ e} > \angle\text{r}$
- ✓
$\angle\text{i} = \angle\text{ e} > \angle\text{r}.$
AnswerCorrect option: D. $\angle\text{i} = \angle\text{ e} > \angle\text{r}.$
When a ray of light passes through a rectangular glass slab then due to change in the refractive index from air to glass and then glass to air, the light is refracted. The angle of incidence is equal to the angle of emergency of light ray and also greater than angle of refraction.
View full question & answer→MCQ 111 Mark
A student used a device $(X)$ to obtain/focus the image of a well illuminated distant building on a screen $(S)$ as shown below in the diagram. Select the correct statement about the device $(X).$
- A
This device is a concave lens of focal length $8 \ cm.$
- B
This device is a convex mirror of focal length $8 \ cm.$
- C
This device is a convex lens of focal length $4 \ cm.$
- ✓
This device is a convex lens of focal length $8 \ cm.$
AnswerCorrect option: D. This device is a convex lens of focal length $8 \ cm.$
The incident rays after passing through the lens converge at the focus. So, the device $'X\ '$ is a converging or a convex lens. The distance between the lens and the screen gives the focal length of the lens.
View full question & answer→MCQ 121 Mark
A student is using a convex lens of focal length $10 \ cm$ to study the image formation by a convex lens for the various positions of the object. In one of his observations, he may observe that when the object is placed at a distance of $20 \ cm$ from the lens, its image is formed at $($select the correct option$)$ :
- A
$20 \ cm$ on the other side of the lens and is of the same size, real and erect.
- B
$40 \ cm$ on the other side of the lens and is magnified, real and inverted.
- ✓
$20 \ cm$ on the other side of the lens and is of the same size, real and inverted.
- D
$20 \ cm$ on the other side of the lens and is of the same size, virtual and erect.
AnswerCorrect option: C. $20 \ cm$ on the other side of the lens and is of the same size, real and inverted.
Given, focal length of the image $, f = 10\ cm.$
Thus, the object is placed at $2F\ (2 \times 10 = 20\ cm)$. Hence, the image is also formed at $2F.$
The image will be of same size as that of the object and is real and inverted.
View full question & answer→MCQ 131 Mark
A student focussed the Sun rays using an optical device $'X\ ’$ on a screen $S$ as shown. From this it may be concluded that the device $'X\ ’ $ is a $($select the correct option$)$
- A
Convex lens of focal length $10\ cm.$
- B
Convex lens of radius of curvature $20\ cm.$
- ✓
Convex lens of focal length $20\ cm.$
- D
Concave mirror of focal length $20\ cm.$
AnswerCorrect option: C. Convex lens of focal length $20\ cm.$
Light rays from infinity are converged at the focal point of the convex lens. Therefore, the focal length of the convex lens is $20\ cm.$
Hence, the correct option is $C$.
View full question & answer→MCQ 141 Mark
A student is using a convex lens of focal length $18 \ cm$ to study the image formation by it for the various positions of the object. He observes that when he places the object at $27 \ cm,$ the location of the image is at $54 \ cm$ on the other side of the lens. Identify from the following diagram the three rays that are obeying the laws of refraction and may be used to draw the corresponding ray diagram
- A
$1, 2$ and $4$
- B
$1, 3$ and $5$
- ✓
$2, 4$ and $5$
- D
$2, 3$ and $4$
AnswerCorrect option: C. $2, 4$ and $5$
Ray $2, 3$ and $4$ are obeys the laws of refraction.
Ray $2$ is parallel to the principal axis and passes through the principal focus after refraction.
Ray $3$ passes from the optical centre of the lens and emerges without any deviation.
Ray $4$ is passing through the principal focus and after refraction from a convex lens emerges parallel to the principal axis.
Ray $1$ and $5$ cannot pass through the focus after refraction as they are not parallel to the principal axis.
Ray $2, 3$ and $4$ are obeys the laws of refraction.
Ray $2$ is parallel to the principal axis and passes through the principal focus after refraction.
Ray $3$ passes from the optical centre of the lens and emerges without any deviation.
Ray $4$ is passing through the principal focus and after refraction from a convex lens emerges parallel to the principal axis.
Ray $1$ and $5$ cannot pass through the focus after refraction as they are not parallel to the principal axis.
View full question & answer→MCQ 151 Mark
A student very cautiously traces the path of a ray through a glass slab for different values of the angle of incidence $(\angle\text{i}).$ He then measures the corresponding values of the angle of refraction $(\angle\text{r})$ and the angle of emergence $(\angle\text{e})$ for every value of the angle of incidence. On analysing these measurements of angles, his conclusion would be
- A
$\angle\text{i} > \angle\text{r} > \angle\text{e}$
- ✓
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
- C
$\angle\text{i} < \angle\text{r} < \angle\text{e}$
- D
$\angle\text{i}=\angle\text{e} < \angle\text{r}$
AnswerCorrect option: B. $\angle\text{i}=\angle\text{e} > \angle\text{r}$
Since the emergent ray is always parallel to the incident ray, therefore the angle of incidence is always equal to the angle of emergence i.e.,
$\angle\text{i}=\angle\text{e}\ .....\text{(i)}$
Also, when a light ray travels from rarer to denser medium, it bends towards the normal. Hence
$\angle\text{i}>\angle\text{r}\ .....\text{(ii)}$
From $(i)$ and $(ii),$ we get
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
Hence, the correct answer is $\angle\text{i}=\angle\text{e} > \angle\text{r}.$
View full question & answer→MCQ 161 Mark
To determine the approximate value of the focal length of a given concave mirror, you focus the image of a distant object formed by the mirror on a screen. The image obtained on the screen, as compared to the object is always :
- A
Laterally inverted and diminished.
- ✓
- C
- D
Erect and highly diminished.
AnswerThe rays from a distant object $($from infinity$)$ always converge at the focus of the mirror. So the position of the screen in this case will give the location of the focus and the distance of screen from the pole of the mirror will give the focal length of the mirror. Also the image formed in such a case is real, inverted and diminished, so the correct answer is real, inverted and diminished.
View full question & answer→MCQ 171 Mark
Suppose you have focused on a screen the image of candle flame placed at the farthest end of the laboratory table using a convex lens. If your teacher suggests you to focus the parallel rays of the sun, reaching your laboratory table, on the same screen, what you are expected to do is to move the :
- ✓
Lens slightly towards the screen.
- B
Lens slightly away from the screen.
- C
Lens slightly towards the sun.
- D
Lens and screen both towards the sun.
AnswerCorrect option: A. Lens slightly towards the screen.
In the case of candle flame, the object is placed beyond $c,$ which means the image is formed or focused between $c$ and $f$. in second case, the object or sun is at infinity, so the image will be formed at focus. This means that the distance between image and lens has decreased in the second case. Either we have to the screen towards the lens or the lens towards the screen.
So, option $(a)$ is correct.
View full question & answer→MCQ 181 Mark
A student traces the path of a ray of light through a triangular glass prism for different values of angle of incidence. On analysing the ray diagrams, which one of the following conclusions is he likely to draw?
- A
The emergent ray is parallel to the incident ray.
- ✓
The emergent ray bends at an angle to the direction of the incident ray.
- C
The emergent ray and the refracted ray are at right angles to each other.
- D
The emergent ray is perpendicular to the incident ray.
AnswerCorrect option: B. The emergent ray bends at an angle to the direction of the incident ray.
In refraction of light through a glass prism, there is deviation or change in the path of light passing through the prism.
View full question & answer→MCQ 191 Mark
A student traces the path of a ray of light passing through a rectangular glass slab and marks the angle of incidence $i,$ angle of refraction $r$ and angle of emergence $e,$ as shown. The correctly marked angle $(s)$ is/are :
AnswerCorrect option: C. $\angle r$ only.
View full question & answer→MCQ 201 Mark
A student has obtained an image of a well-illuminated distant object on a screen to determine the focal length, $\text{F}_{1}$ of the given spherical mirror. The teacher then gave him another mirror of focal length, $\text{F}_{2}$ and asked him to obtain a focussed image of the same object on the same screen. The student found that in order to focus the same object using the second mirror, he has to move the mirror away from the screen. From this observation it may be concluded that both the spherical mirrors given to the student were $($select the correct option$)$
- ✓
Concave and $\text{F}_{1} < \text{F}_{2}$
- B
Concave and $\text{F}_{1} > \text{F}_{2}$
- C
Convex and $\text{F}_{1} < \text{F}_{2}$
- D
Convex and $\text{F}_{1} > \text{F}_{2}$
AnswerCorrect option: A. Concave and $\text{F}_{1} < \text{F}_{2}$
Since the image is focused, the spherical mirror is a concave mirror. For second mirror the distance is increased to focus the image on the screen. Hence, focal length is more than that of first mirror.
View full question & answer→MCQ 211 Mark
A student focussed the Sun rays using an optical device $'X\ ’$ on a screen $S$ as shown. From this it may be concluded that the device $'X\ ’$ is a $($select the correct option$)$
- A
Convex lens of focal length $10 \ cm.$
- B
Convex lens of radius of curvature $20 \ cm.$
- ✓
Convex lens of focal length $20 \ cm.$
- D
Concave mirror of focal length $20 \ cm.$
AnswerCorrect option: C. Convex lens of focal length $20 \ cm.$
Light rays from infinity are converged at the focal point of the convex lens. Therefore, the focal length of the convex lens is $20 \ cm.$
Hence, the correct option is $C.$
View full question & answer→MCQ 221 Mark
He then measures the corresponding values of the angle of $(\angle\text{i}).$A student very cautiously traces the path of a ray through a glass slab for different values of the angle of incidence refraction $(\angle\text{r}).$ and the angle of emergence $(\angle\text{e}).$ for every value of the angle of incidence. On analysing these measurements of angles, his conclusion would be.
- A
$\angle\text{i} > \angle\text{r} > \angle\text{e}$
- ✓
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
- C
$\angle\text{i} < \angle\text{r} < \angle\text{e}$
- D
$\angle\text{i}=\angle\text{e} < \angle\text{r}$
AnswerCorrect option: B. $\angle\text{i}=\angle\text{e} > \angle\text{r}$
Since the emergent ray is always parallel to the incident ray, therefore the angle of incidence is always equal to the angle of emergence i.e.,
$\angle\text{i}=\angle\text{e}\ .....\text{(i)}$
Also, when a light ray travels from rarer to denser medium, it bends towards the normal. Hence
$\angle\text{i} > \angle\text{r}\ .....\text{(ii)}$
From $(i)$ and $(ii),$ we get
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
Hence, the correct answer is $\angle\text{i}=\angle\text{e}>\angle\text{r}.$
View full question & answer→MCQ 231 Mark
A student used a device $(X)$ to obtain/ focus the image of a well illuminated distant building on a screen $(S)$ as shown below in the diagram. Select the correct statement about the device $(X).$
- A
This device is a concave lens of focal length $8\ cm.$
- B
This device is a convex mirror of focal length $8\ cm.$
- C
This device is a convex lens of focal length $4\ cm.$
- ✓
This device is a convex lens of focal length $8\ cm.$
AnswerCorrect option: D. This device is a convex lens of focal length $8\ cm.$
The incident rays after passing through the lens converge at the focus. So, the device $'X\ '$ is a converging or a convex lens. The distance between the lens and the screen gives the focal length of the lens.
View full question & answer→MCQ 241 Mark
A student obtains a sharp image of the distant window $(W)$ of the school laboratory on the screen $(S)$ using the given concave mirror $(M)$ to determine its focal length. Which of the following distances should he measure to get the focal length of the mirror?
- A
$MW.$
- ✓
$MS.$
- C
$SW.$
- D
$MW – MS.$
AnswerTo get the focal length of the mirror, the student should measure the distance between mirror and screen because when the object is at infinity, the image is formed at focus i.e. between screen and mirror. So, the distance between screen and mirror should be measured.
View full question & answer→MCQ 251 Mark
After tracing the path of a ray of light passing through a rectangular glass slab for four different values of the angle of incidence, a student reported his observations in tabular form as given below :
| S. No. |
$\angle\text{ i}$ |
$\angle\text{ r}$ |
$\angle\text{ e}$ |
| $I$ |
$30^{0}$ |
$19^{0}$ |
$29^{0}$ |
| $II$ |
$40^{0}$ |
$28^{0}$ |
$40^{0}$ |
| $III$ |
$50^{0}$ |
$36^{0}$ |
$50^{0}$ |
| $IV$ |
$60^{0}$ |
$40^{0}$ |
$59^{0}$ |
The best observation is : AnswerFor best results, the following conditions must be satisfied :
- The angle of incidence must be nearly equal to the angle of emergence.
- The angle of refraction must be less than the angle the incidence.
- $\frac{\sin\text{i}}{\sin\text{r}}=1.5\ ($Since, the refractive index of glass is $1.5)$
Out of the $4$ observations, the first observation, that is angle $i = 30^\circ ;$ angle $r = 19^\circ $ and angle $e = 29^\circ $ is the correct observation. In this case, $\frac{\sin 30^\circ}{\sin 19^\circ}=1.535$. View full question & answer→MCQ 261 Mark
A student is using a convex lens of focal length $10 \ cm$ to study the image formation by a convex lens for the various positions of the object. In one of his observations, he may observe that when the object is placed at a distance of $20 \ cm$ from the lens, its image is formed at $($select the correct option$) :$
- A
$20\ cm$ on the other side of the lens and is of the same size, real and erect.
- B
$40\ cm$ on the other side of the lens and is magnified, real and inverted.
- ✓
$20\ cm$ on the other side of the lens and is of the same size, real and inverted.
- D
$20\ cm$ on the other side of the lens and is of the same size, virtual and erect.
AnswerCorrect option: C. $20\ cm$ on the other side of the lens and is of the same size, real and inverted.
$20\ cm$ on the other side of the lens and is of the same size, real and inverted.
Given, focal length of the image $, f = 10\ cm.$
Thus, the object is placed at $2F\ (2 \times 10 = 20\ cm)$. Hence, the image is also formed at $2F.$
The image will be of same size as that of the object and is real and inverted.
View full question & answer→MCQ 271 Mark
A student traces the path of a ray of light through a rectangular glass slab for four different angles of incidence. He very cautiously measures the angle $f,$ angle $r$ and the angle $e$. On analysing his measurements, he is likely to draw the following conclusion :
- ✓
$\angle\text{i} = \angle\text{e} > \angle\text{r}$
- B
$\angle\text{i} > \angle\text{r} > \angle\text{e}$
- C
$\angle\text{i} = \text{Zr} < \text{Ze}$
- D
$\angle\text{i} = \angle\text{e}< \angle\text{r}$
AnswerCorrect option: A. $\angle\text{i} = \angle\text{e} > \angle\text{r}$
Since the emergent ray is always parallel to the incident ray, therefore the angle of incidence is always equal to the angle of emergence i.e.
$\angle\text{i}=\angle\text{e}\ .....(\text{i})$
Also, when a light ray travels from rarer to denser medium, it bends towards the normal. Hence
$\angle\text{i}>\angle\text{r}\ .....(\text{ii})$
From $(i)$ and $(ii),$ we get
$\angle\text{i}=\angle\text{e} > \angle\text{r}$
Hence, the correct answer is $\angle\text{i}=\angle\text{e} > \angle\text{r}$
View full question & answer→MCQ 281 Mark
A student has to determine the focal length of a concave mirror by obtaining the image of a distant object on a screen. For getting best result he should focus :
AnswerCorrect option: A. A distant tree or an electric pole.
View full question & answer→MCQ 291 Mark
If you are to determine to focal length of a convex lens, you should have
- A
A convex lens and a screen.
- B
A convex lens and a lens holder.
- C
A lens holder, a screen holder and a scale.
- ✓
A convex lens, a screen, holder for them and a scale.
AnswerCorrect option: D. A convex lens, a screen, holder for them and a scale.
View full question & answer→MCQ 301 Mark
While performing the experiment on tracing the path of a ray of light through a rectangular glass slab, in which of the following experimental set $-$ ups is a student likely to get best results $?\ P_1$ and $P_2$ are the positions of pins fixed by him.
View full question & answer→MCQ 311 Mark
Four students showed the following traces of the path of a ray of light passing through a rectangular glass slab. The trace most likely to be correct is that of student :
MCQ 321 Mark
A student obtained a sharp inverted image of a distant tree on a screen placed in front of the concave mirror. He then removed the screen and tried to look into the mirror. He would now see
- A
A very blurred image on the wall opposite to the mirror.
- B
An erect and magnified image of the tree in the mirror.
- C
No image as the screen has been removed.
- ✓
A highly diminished inverted image of the tree at the focus of the mirror.
AnswerCorrect option: D. A highly diminished inverted image of the tree at the focus of the mirror.
View full question & answer→MCQ 331 Mark
The steps involved in observing a slide under a microscope are given below. They may not be in proper sequence.
- Focus the object under high power of the microscope.
- Place the slide on the stage of the microscope.
- Arrange the mirror to reflect maximum light to the slide.
- Focus the object under low power of the microscope.
The proper sequence of steps is:
- ✓
$\ce{II, III, IV, I}$
- B
$\ce{I, II, III, IV}$
- C
$\ce{IV, III, II, I}$
- D
$\ce{III, I, II, IV}$
AnswerCorrect option: A. $\ce{II, III, IV, I}$
View full question & answer→MCQ 341 Mark
To determine the focal length of a convex lens by obtaining a sharp image of a distant object, the following steps were suggested which are not in proper sequence :
- Hold the lens between the object and the screen.
- Adjust the position of the lens to form sharp image.
- Select a suitable distant object.
- Measure the distance between the lens and the screen.
The correct sequence of steps to determine the focal length of the lens is :
- ✓
$\ce{III, I, II, IV}$
- B
$\ce{III, I, IV, II}$
- C
$\ce{III, IV, II, I}$
- D
$\ce{I, II, III, IV}$
AnswerCorrect option: A. $\ce{III, I, II, IV}$
View full question & answer→MCQ 351 Mark
To find the focal length of a concave mirror, Sita should choose which one of the following set $-$ ups?
- A
A mirror holder and a screen holder.
- B
A screen holder and a scale.
- C
A mirror holder, a screen holder and a scale.
- ✓
A screen, a mirror, holders for them and a scale.
AnswerCorrect option: D. A screen, a mirror, holders for them and a scale.
View full question & answer→MCQ 361 Mark
For a ray of light passing through a glass slab. The lateral displacement was correctly measured as :
MCQ 371 Mark
In these diagrams, the angle of refraction r has been correctly marked in which diagram?
MCQ 381 Mark
By using convex lens, a student obtained a sharp image of his class $-$ room window grill on a screen. In which direction should he move the lens to "focus a distant tree instead of the grill?
- ✓
- B
- C
Very far away from the screen.
- D
View full question & answer→MCQ 391 Mark
When you focus the image of a distant flag, whose shape is given below, on a screen using a convex lens, the shape of the image as it appears on the screen is:
MCQ 401 Mark
An optical device has been given to a student and he determines its focal length by focusing the image of the sun on a screen placed $24\ cm$ from the device on the same side as the sun. Select the correct statement about the device.
- A
Convex mirror of focal length $12\ cm.$
- B
Convex lens of focal length $24\ cm.$
- ✓
Concave mirror of focal length $24\ cm.$
- D
Convex lens of focal length $12\ cm.$
AnswerCorrect option: C. Concave mirror of focal length $24\ cm.$
View full question & answer→MCQ 411 Mark
In an experiment to trace the path of a ray of light through a triangular glass prism, a student would observe that the emergent ray.
AnswerCorrect option: D. Gets deviated and bends towards the thicker part $($base$)$ of the prism.
If the refractive index of the medium $($dispersing medium in our case "glass"$)$ through which the light passes during dispersion is greater than the medium through which the light enters $($in our case "air"$)$ the dispersing medium, then the net deviation in the path of light is always towards the thicker part of the prism.
View full question & answer→MCQ 421 Mark
An optical device has been given to a student and he determines its focal length by focusing the image of the sun on a screen placed $24 \ cm$ from the device on the same side as the sun. Select the correct statement about the device.
- A
Convex mirror of focal length $12 \ cm.$
- B
Convex lens of focal length $24 \ cm.$
- ✓
Concave mirror of focal length $24 \ cm.$
- D
Convex lens of focal length $12 \ cm.$
AnswerCorrect option: C. Concave mirror of focal length $24 \ cm.$
View full question & answer→MCQ 431 Mark
A student has traced the path of a ray of light through a glass slab as follows. If you are asked to label $\{1, 2, 3\}$ and $4,$ the correct sequencing of labeling $\angle\text{i}, \angle\text{e}, \angle\text{r}$ and lateral displacement respectively is
- A
$\{2, 1, 3, 4\}$
- B
$\{1, 2, 3, 4\}$
- ✓
$\{1, 3, 2, 4\}$
- D
$\{1, 3, 4, 2\}$
AnswerCorrect option: C. $\{1, 3, 2, 4\}$
View full question & answer→MCQ 441 Mark
A student determines the focal length of a device $'X\ ’$ by focusing the image of a distant object on a screen placed $20\ cm$ from the device on the same side as the object. The device $'X\ ’$ is.
- A
Concave lens of focal length $10\ cm.$
- B
Convex lens of focal length $20\ cm.$
- C
Concave mirror of focal length $10\ cm.$
- ✓
Concave mirror of focal length $20\ cm.$
AnswerCorrect option: D. Concave mirror of focal length $20\ cm.$
Since the object and the image are on the same side, and the image is real $($since it is focused on a screen$)$ the device $X$ is a concave mirror of focal length $20\ cm.$
View full question & answer→MCQ 451 Mark
A teacher sets up the stand carrying a convex lens of focal length $15 \ cm$ at $42.7 \ cm$ mark on the optical bench. He asks four students $\ce{A, B, C}$ and $D$ to suggest the position of screen on the optical bench so that a distinct image of a distant tree is obtained almost immediately on it. The positions suggested by the students were as:
- $12.7 \ cm.$
- $29.7 \ cm.$
- $57.7 \ cm.$
- $72.7 \ cm.$
The correct position of the screen was suggested by.
AnswerA ray comes from infinity, after refraction through a convex lens, meet at the second principal focus. So, screen should be placed at the second principal focus of a convex lens. e.g., screen should be placed at $15\ cm$ from the pole of convex lens.
Now, the position of the screen on the optical bench $=$ distance between screen from the pole of convex lens $+$ distance of optical bench from the pole of convex lens as you can see in attachment
$= 15 \ cm + 42.7\ cm$
$= 57.7\ cm$
Hence, option $(C)$ is correct
View full question & answer→MCQ 461 Mark
A teacher sets up the stand carrying a convex lens of focal length $15\ cm$ at $42.7\ cm$ mark on the optical bench. He asks four students $\ce{A, B, C} $ and $D$ to suggest the position of screen on the optical bench so that a distinct image of a distant tree is obtained almost immediately on it. The positions suggested by the students were as:
- $12.7\ cm.$
- $29.7\ cm.$
- $57.7\ cm.$
- $72.7\ cm.$
The correct position of the screen was suggested by.
AnswerA ray comes from infinity, after refraction through a convex lens, meet at the second principal focus. So, screen should be placed at the second principal focus of a convex lens. e.g., screen should be placed at $15\ cm $ from the pole of convex lens.
Now, the position of the screen on the optical bench $=$ distance between screen from the pole of convex lens $+$ distance of optical bench from the pole of convex lens as you can see in attachment
$= 15\ cm + 42.7\ cm$
$= 57.7\ cm$
Hence, option $(C)$ is correct.
View full question & answer→MCQ 471 Mark
In the following diagram the correctly marked angles are :
- ✓
$\angle A$ and $\angle e$
- B
$\angle i, \angle A$ and $\angle D$
- C
$\angle A, \angle r $ and $\angle e$
- D
$\angle A, \angle r$ and $\angle D$
AnswerCorrect option: A. $\angle A$ and $\angle e$
The correctly marked angles in the given figure are the angle of prism $(A)$ and the angle of incidence $(i).$
All the angles are correctly marked in the figure shown below.
Hence, the correct answer is option $(a).$ View full question & answer→MCQ 481 Mark
If you focus the image of a distant object, whose shape is given below, on a screen using a convex lens,
the shape of the image of this object on the screen would be:
MCQ 491 Mark
Study the following figure in which a student has marked the angle of incidence (∠i), angle of refraction (∠r), angle of emergence (∠e), angle of prism (∠A) and the angle of deviation (∠D). The correctly marked angles are:
MCQ 501 Mark
Four students $P, Q, R$ and $S$ traced the path of a ray of light passing through a glass slab for an angle of incidence of $40^\circ $ and measured the angle of refraction. The values as measured by them were $18^\circ ; 22^\circ ; 25^\circ $ and $30^\circ $ respectively. The student who has performed the experiment methodically is.
AnswerSince, angle of incidence, $\theta_1=40^\circ$ and we have to find angle of emergence, $\theta_2.$
Since $,\frac{\sin\theta_1}{\sin\theta_2}=\frac{\text{n}_2}{\text{n}_1}$
Refractive index of air, $\mathrm{n}_1=1$
Refractive index of glass, $\mathrm{n}_2=1.5$
$\frac{\sin 40^\circ}{\sin \theta_2}=\frac{1.5}{1}$
$\sin\theta_2=\sin 40^\circ\times\frac{1}{1.5}$
$\sin \theta_2=\frac{0.642}{1.5}=0.428$
Now, $\theta_2=\sin^{-1}(0.428)\approx25^\circ$
View full question & answer→
