Water in a concave mirror?
Wow...that's nasty. Either it's a trick question (somehow the answer's the same as if there were no water, although I doubt it), or you have to assume that the mirror is small enough that the entire surface is covered by water (else you'd get two images...one from the covered part and one from the uncovered part). With the limitation of 1cm max depth, that means the mirror is about 20cm across. (A triangle of 50cm hypotenuse and a long right angle leg of 49cm in air means the short right angle leg is 9.95cm). On that mirror without any water, object distance equals image distance when the object is at twice the focal length, or 100cm in this case. The water results in a perturbation from that, so we can probably safely assume small angle, as all angles will be less than 6° away from normal when hitting the water interface. Let's assume a liquid with really high index of refraction. At some point, increasing n will make it so that you get total internal reflection at the larger angles, so you need to back off farther to avoid this. Also, you're gonna get more spherical aberration from the water layer, so it may not even be possible to get a focused image if you work things out in gory detail. Unless there's something in your lecture notes or textbook that gives a trick way around this, I'm gonna have to call it at either "100cm" or "no image" depending on whether you want to play loose approximations or full detail.
Conceptual questions about mirrors?
1. The image is not inverted when the image is virtual. This occurs when the object is between the mirror surface and the focal point. 2. The magnified image will occupy a larger area than the original object, and the amount of light being scattered from the original object will be spread over a larger area. This results in a lower power-density for the magnified image. 3. Real images will be inverted, virtual images are not inverted.
What are the viva questions asked for physics practical?
There are many website that can help some of them are mentioned below :1) Questions asked in Physics Viva Voce for Class 12 Physics Practical » CBSE Physics2) An Exemplary Viva Voce (CBSE Class XII Physics Practical)3) CBSE Class XII Physics and Chemistry Viva Questions Class XII Science4) CBSE Class 12 Physics practical exam tips and tricksHope this will be helpful. All the Best !
Is it possible for a convex mirror to produce a real image when the object is virtual?
Yes, it is.The trick is to produce a virtual object. While a real object produces divergent rays (like a source), a virtual object can be something that receives convergent rays (like a sink). If we agree on this, then it's easy to form real images using convex mirrors (with a separate arrangement for producing convergent rays from real objects).
(NO CALCULATION REQUIRED) WHAT IS THE ANSWER FOR THIS PHYSICS QUESTION?
An object is placed to the left of a spherical mirror in front of the mirror. If the image of the object is formed on the right side of the mirror, which of the following statements must be true? (There could be more than one correct choice.) A) The image is inverted and real, and the mirror must be convex. B) The mirror could be either concave or convex. C) The image is upright and the mirror must be convex. D) The image is upright and the mirror must be concave. E) The image is upright and virtual.
Is it possible that a lens acts as a convergent lens in one medium and divergent lens in another medium?
Yes, any refraction that happens is due to the difference in densities of the medium w.r.t speed of light in each.Moreover, refractive index is a relative term meaning one medium is denser/rarer than the other medium is based on the other medium entirely. Write in comments if your need further explanation regarding mathematical relations.
Can a concave lens behave as a convex lens by placing it in a different medium?
Concave lens can behave as a convex lens when placed in a medium whose refractive index is greater than that of the material of which the concave lens is made.This is because the focal length of a concave lens is directly proportional to the refractive index of the medium (in which it is kept) i.e.1/f= (n–1)(1/R1 - 1/R2), where n is the refractive index (or index of refraction), and n=n(lens)/n(medium)and so, focal length of the concave lens will increase from a negative value to a positive value hence, power of this lens would also become positive (as focal length of this lens is now positive i.e. P =1/f) thus, it will behave as a convex lens in another medium (whose refractive index is greater than that of the material of which the concave lens is made).
Rank the images on their bases of size largest to smallest (physics).?
If d is the distance of the object from the mirror and i is the distance of the image from the mirror, then the magnification m = -i/d, the negative sign being mathematics shorthand for the inversion of the image. Using the equation 1/d + 1/i = 1/f, we can solve for i to get i = df / (d − f) and m = -f / (d − f) if d > f, i.e., d − f > 0, then the image is a real image in front of the mirror and can be projected onto a screen, provided that the object and/or screen doesn't get in the way of the light path. Otherwise, the image is a virtual one that appears behind the mirror and that requires an auxiliary optic, e.g., the cornea and lens of an eye, to form an image. I get magnifications of 1) 4x (virtual) 2) -1x (real) 3) -1x (real) 4) 2x (virtual) 5) -0.5x (real) 6) 4/3x (virtual)
Rank the images on their bases of size largest to smallest (physics).?
Have two equations: (1/f) = (1/do)+(1/di) where f is the focal length, do is the object distance, di is the image distance and hi = -hodi/do where hi is the image height and ho is the object height 1. (1/20)=(1/15)+(1/di) (1/di) = -1/60 di=-60 hi = 60ho/15=4ho 2. (1/5)=(1/10)+(1/di) 1/di = 1/10 di=10 hi= -10ho/10 = 10 = -ho 3. (1/10)=(1/20)+(1/di) 1/di = 1/20 di=20 hi=-20ho/20 = -ho 4. (1/20)=(1/10)+(1/di) 1/di = -1/20 di=-20 hi=20ho/10 = 2ho 5. (1/5)=(1/15)+(1/di) 1/di = 2/15 di = 15/2 hi = -15ho/2*15 = -ho/2 6. (1/20) = (1/5) + (1/di) 1/di = -3/20 di = -20/3 hi = 20ho/3*5 = 4ho/3 So in order of size we have: 1,4,6,(2,3=),5