00:01
Hello everyone here in this question we are given a concave mirror.
00:04
It is given that the radius of curvature of the mirror r is equal to minus 10 cm.
00:09
We have taken minus with the radius of curvature with respect to the sign convention.
00:14
The height of an object which is a candle here is given as 5 cm.
00:18
The object distance u is equal to minus 15 cm.
00:21
We have taken minus with 15 cm because it is placed in front of the mirror.
00:26
We are required to find out the image distance we take the image distance to be v we're also required to find out the height of the image that is formed we consider this height to be h now first of all let us find out the focal length of the mirror the focal length f is given by the radius of curvature are divided by two so from here we'll have the focal length f is equal to the radius of curvature is minus 10 centimeter so minus 10 divided by 2 which is equal to minus 5 cm now to find out the image distance will apply the mirror equation according to this mirror equation we have 1 divided by the object distance plus 1 divided by the image distance is equal to 1 divided by the focal length so we get 1 divided by the image distance is equal to 1 divide by the focal length minus 1 divided by the object distance now putting the values 1 divided by the focal link is minus 5 cm minus 1 divided by the object distance is minus 15 centimeter now this gives us minus 2 by 15 upon further simplifying the image distance v is equal to minus 15 by 2 which is equal to minus 7 .5 centimeter now we find out the height of the image for that we'll use magnification of the image m which is given by negative of the image distance divided by the object distance which is also equal to the height of image divided by the height of the object supporting the values will get negative of the image distance which is minus 7 .5 cm divided by the object distance which is given as minus 15 cm this is equal to the height of image divided by the height of object which is mentioned as 5 cm from here we will get the height h which is equal to negative of 2 .5 centimeter the negative sign here implies that the image is an inverted one therefore the image distance is minus 7 .5 centimeter and its height is minus 2 .5 cm.
03:26
Hence we have the answer of the first part of the question.
03:29
Now we move on to the next part.
03:32
The second part we are given that array of light travels from liquid to glass.
03:36
The refractive index of liquid n1 is equal to 1 .75 and the refractive index of glass n2 is equal to 1 .52.
03:44
We are given the angle of incidence i which is equal to 59 degree.
03:48
We are required to find out if the light will refract or to undergo total internal reflection.
03:55
To solve this question, we'll apply first snell's law.
03:59
According to snell's law, sign of angle of incidence i divided by sign of angle of refraction r is equal to the refractive index of the medium in which the refraction is happening.
04:12
Here it is glass, so n2 divided by the repetitive index of the incident medium that is n1...