Explanation:
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.
Explanation:
Since, 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, n1 = 1
Refractive index of glass, n2 = 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, $\theta_2=\sin^{-1}(0.428)\approx25^\circ$
Explanation:
From the figure, we can see that the light rays coming from infinity get reflected by the device 'X' and converge at a point at a distance 12cm from it. Therefore, the device 'X' is a concave mirror of focal length 12cm.
Explanation:
The lens is convex, as it forms real image. As mentioned in the second case, the image distance (v) is increasing; hence, the object distance (u) is decreasing. The lens formula is 1/ F = 1/ v - 1/ u ⇒ F = uv/ u - v For convex lens, object distance = -u Image distance = +v ⇒ F = uv/ u + v where F is the focal length of the lens. Therefore, F is lesser for the second lens of focal length F2. Hence, F1 > F2.
Hence, the correct option is C.
Explanation:
Angle 2 is angle of incidence. As, it is formed between the incident ray and the normal.
Angle 1 is angle of emergence. As, it is formed between the emergent ray with normal.
Angle 4 is angle of refraction. As, it is formed between the refracted ray and the normal.
3 shows the lateral displacement.
Hence, the correct answer is 2, 1, 4, 3
Explanation:
When the object is at infinity, then the image is formed at the focus of the concave mirror. The image formed in this case is always inverted and diminished i.e. smaller in size as that of the object.
Explanation:
Experiment II will give the best result because it has the largest angle of incidence, due to which the lateral displacement between the incident ray and emergent ray will be maximum.
Explanation:
Convex lens is a converging lens. Therefore, rays I and III represent the path of the refracting ray from a convex lens (converging lens). Figure II and IV show the refracting ray being diverged. So, the ray diagrams are incorrect.
S. No. | Object distance(cm) | Image distance (cm) |
1 | 60 | 15 |
2 | 48 | 16 |
3 | 36 | 21 |
4 | 24 | 24 |
5 | 18 | 36 |
6 | 16 | 48 |
Explanation:
In observation No. 3 the focal length comes out to be 13.26cm whereas, the focal length for all other observation is 12cm.
Explanation:
To measure the focal length of the mirror, the object should be taken at infinity. Therefore, the image formed by the concave mirror would be real, inverted, diminished and red in shade.
Hence, the correct option is C.
Explanation:
A concave mirror always forms the image of a distant object at its focus. In the given case the distance between the mirror and the screen is equal to the focal length of the mirror. Thus, the student A measures the focal length correctly.
Explanation:
The object, i.e., the distant building is at a distance of 1km from the laboratory. That is, for the concave mirror, the building (object) is at infinity. In order to obtain a sharp image of the building, the student must slightly move the screen towards the mirror.
Explanation:
The lens is convex, as it forms real image. As mentioned in the second case, the image distance (v) is increasing; hence, the object distance (u) is decreasing. The lens formula is 1/ F=1/ v-1/ u ⇒ F = uv/ u-v For convex lens, object distance = -u Image distance = +v ⇒ F = uv/ u + v where F is the focal length of the lens. Therefore, F is lesser for the second lens of focal length F2. Hence, F1 > F2 Hence, the correct option is C.
Explanation:
The paths followed by the incident ray and the emergent ray on passing through a prism are different. Upon refraction of a ray of light through prism, the emergent ray is always at an angle (namely, the Angle of Deviation) with the incident ray.
Thus, option (4) is correct.
Explanation:
As the subject teacher suggested the student for focusing a will illuminated distant object, the image distance is increased. Due to this the object distance would decrease and thus the distance between the mirror and the screen should be decreased. Therefore, the mirror should be moved slightly towards the screen.
| S. No. | $\angle\text{ i}$ | $\angle\text{ i}$ | $\angle\theta$ |
| I | $30^{0}$ | $20^{0}$ | $31^{0}$ |
| II | $40^{0}$ | $25^{0}$ | $40^{0}$ |
| III | $50^{0}$ | $31^{0}$ | $49^{0}$ |
Explanation:
In a glass slab refraction: angle of incidence (i) = angle of emergence (e) Since here in all cases angle of incidence (i) is nearly equal to the angle of emergence (e), so all the values are correct.
S. No. | $\angle \text{ i}$ | $\angle \text{ r}$ | $\angle \text{ e}$ |
I | $60^{0}$ | $40^{0}$ | $61^{0}$ |
II | $50^{0}$ | $36^{0}$ | $51^{0}$ |
III | $40^{0}$ | $28^{0}$ | $39^{0}$ |
IV | $30^{0}$ | $20^{0}$ | $31^{0}$ |
Explanation:
The correct sequence to find the focal length of a convex lens for obtaining a sharp image of a distant object will be as follows: Select a well lit distant object, Place a screen opposite to the object on the lab table, Adjust the position of the lens to form a sharp image. Hold the lens between the object and the screen and they Measure the distance between the lens and the screen.
Explanation:
When the image distance increases, object distance decreases. Thus, distance between the mirror and screen will decrease. So, the mirror should be moved towards the screen.
Incident and emergent angles are always measured from the normal of the plane.
Explanation:
The chosen object must be sun, because parallel rays coming from infinity converge or meet at focus.
| S. No. | Angle of incidence | Angle of refraction | Angle of emergence |
| I | $45^{0}$ | $41^{0}$ | $45^{0}$ |
| II | $40^{0}$ | $38^{0}$ | $38^{0}$ |
| III | $45^{0}$ | $41^{0}$ | $40^{0}$ |
| IV | $41^{0}$ | $45^{0}$ | $41^{0}$ |
Explanation:
Correct option is an inverted image of the object at the focus of the lens. We use screen just to show the presence of the image but if we remove it image will still forms at focus only.
Explanation:
A well lit distant tree will be best to use as the brighter object will give more accurate and brighter image of the object. In this case it'll be easy for us to find the focal length of the lens.
Explanation:
We know that from the laws of reflection, the incident ray, the reflected ray and the normal to the reflecting surface all lie in the same plane.
Also, the angle of reflection is equal to the angle of incidence.
The laws of reflection hold good for all reflecting surfaces irrespective of their shapes whether plane or curved.
Explanation:
Object between Principal Focus (F) and Pole (P): When the object is placed between principal focus and pole of a concave mirror, an enlarged, virtual and erect image is formed behind the mirror.
Explanation:
The given material having their refractive index as kerosene is 1.44, water is 1.33, musterd oil is 1.46 and glycerine is 1.74. Thus, glycerine is most optically denser and hence have the largest refractive index. Therefore, ray of light bend most in glycerine.
Explanation: The positive sign with the focal length indicates that it is a convex lens.
Explanation:
The diverging lens is signified by negative sign. Also, more is the magnitude of the lens more is the more is the diverging power.
Explanation: When a point source of light is at the focus of a concave mirror or a convex lens, then emergent rays of light make a parallel beam of light. This is the reason; concave mirror is used as reflector of headlights.
Explanation:
When an object is placed at the focus of a concave mirror (convex lens), the reflected (refracted) light rays are always parallel to each other and to the principal axis.
Explanation:
A concave lens always forms a virtual image, smaller than the size of the object.
Explanation:
Concave mirror to be used ti obtain a parallel beam of light from a small lamp.
Explanation:
Both a concave mirror and a convex lens focus parallel light beams coming from a distant object onto the focus.
Explanation:
Because if a ray of light goes form a denser medium (glass) to a rarer medium (water), it bends away from the normal.
Explanation:
A concave lens always forms a virtual, erect and diminished image.
Explanation:
It should be a convex mirror. This is because when an object is in front of a convex mirror, irrespective of its distance, a virtual, erect and diminished image of the object is obtained.
Explanation:
A concave mirror can form an image larger than the actual object when the object is placed between the focus and centre of curvature. Image formation is shown below, where object (AB) is placed between the focus and centre of curvature for which image (A'B') is obtained behind the mirror.
Explanation:
Image formed by a plane mirror is virtual, erect and of the same size as the object.Image formed by a concave mirror is virtual, erect and larger than the object.
