Correct answer Carries: 4.
Wrong Answer Carries: -1.
When light travels from a denser medium to a rarer medium and the angle of incidence exceeds a certain value, what phenomenon occurs?
When light travels from a denser to a rarer medium (e.g., glass to air) and the angle of incidence exceeds the critical angle, total internal reflection occurs. This is because the refracted ray would otherwise require a sine value greater than 1, which is physically impossible.
A ray of light passes from water (\( n = 1.33 \)) to glass (\( n = 1.62 \)) at an angle of incidence of \( 45^\circ \). What is the angle of refraction?
Snell’s law: \( n_1 \sin i = n_2 \sin r \).
Water (\( n_1 = 1.33 \)), glass (\( n_2 = 1.62 \)), \( i = 45^\circ \).
\( 1.33 \times \sin 45^\circ = 1.62 \times \sin r \).
\( \sin 45^\circ = 0.707 \Rightarrow 1.33 \times 0.707 = 1.62 \sin r \Rightarrow 0.941 = 1.62 \sin r \).
\( \sin r = \frac{0.941}{1.62} \approx 0.581 \Rightarrow r = \sin^{-1}(0.581) \approx 35.5^\circ \).
What happens to the image formed by a convex mirror if the object is moved closer to the mirror from a distant position?
A convex mirror always forms a virtual, erect, and diminished image. As the object moves closer, the image size increases slightly but remains diminished (less than the object size), and the image distance increases, approaching the focal length as a limit, though it never exceeds it.
In a compound microscope, why is the final image inverted with respect to the object?
In a compound microscope, the objective lens forms a real, inverted image of the object. The eyepiece then acts as a magnifying lens, forming a virtual image of this inverted intermediate image. Since the eyepiece does not reinvert the image, the final image remains inverted relative to the original object.
A convex lens (\( f = 60 \, \text{cm} \)) and a concave lens (\( f = 30 \, \text{cm} \)) are in contact. What is the effective focal length?
\( f_1 = 60 \, \text{cm} \), \( f_2 = -30 \, \text{cm} \).
\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60} + \frac{1}{-30} = \frac{1}{60} - \frac{2}{60} = \frac{1 - 2}{60} = \frac{-1}{60} \).
\( f = -60 \, \text{cm} \) (diverging system).
An object is placed \( 12 \, \text{cm} \) from a convex mirror of focal length \( 20 \, \text{cm} \). What is the magnification?
\( f = 20 \, \text{cm} \), \( u = -12 \, \text{cm} \).
\( \frac{1}{v} + \frac{1}{-12} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3 + 5}{60} = \frac{8}{60} = \frac{2}{15} \).
\( v = 7.5 \, \text{cm} \).
Magnification: \( m = -\frac{v}{u} = -\frac{7.5}{-12} = 0.625 \).
In a concave lens, why is the image always formed on the same side as the object?
A concave lens diverges light rays, making them appear to originate from a point on the same side as the object when traced backward. This results in a virtual image that cannot be projected on a screen, always forming on the object’s side regardless of its position.
A concave lens of focal length \( 10 \, \text{cm} \) forms an image \( 5 \, \text{cm} \) from the lens. What is the object distance?
Focal length: \( f = -10 \, \text{cm} \) (concave lens).
Image distance: \( v = -5 \, \text{cm} \) (virtual image).
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
\( \frac{1}{-5} - \frac{1}{u} = \frac{1}{-10} \Rightarrow \frac{1}{u} = \frac{1}{-5} - \frac{1}{-10} = \frac{-2 + 1}{10} = \frac{-1}{10} \).
\( u = -10 \, \text{cm} \).
In a convex lens, if an object is placed at the focal point, where is the image formed?
For a convex lens, when the object is at the focal point (F), the rays after refraction become parallel and do not converge to a point on the other side. The image is formed at infinity, as the rays appear to diverge from an infinitely distant point when traced backward.
Why does the apparent depth of an object in a denser medium appear less than its real depth when viewed from air?
When light travels from a denser medium (e.g., water) to a rarer medium (air), it bends away from the normal. This refraction makes the rays appear to diverge from a point closer to the surface than the actual object, reducing the apparent depth compared to the real depth.
Are you sure you want to submit your answers?