An object is placed at 30.0 cm in front of a concave mirror of focal length 10.0 cm then image distance is

1. An object is placed 10 cm from a of image Known : The focal length (f) = 5 cm The object distance ( d o ) = 10 cm Solution : Formation of image by concave mirror : The image distance : 1/ d i = 1/f – 1/ d o = 1/5 – 1/10 = 2/10 – 1/10 = 1/10 d i = 10/1 = 10 cm The image distance is 10 cm. The magnification : m = – d i / d o = -10/10 = -1 1 means that the image is the same as the object. The minus sign indicates that the image is inverted. If the sign is positive than the image is upright. [irp] 2. A 5-cm-high object is placed in front of a concave mirror with a radius of curvature of 20 cm. Determine the image height if the object distance is 5 cm, 15 cm, 20 cm, 30 cm. Known : The radius of curvature ( r ) = 20 cm The focal length (f) = R/2 = 20/2 = 10 cm The object height (h o ) = 5 cm Solution : a) the focal length (f) = 10 cm and the object distance ( d o ) = 5 cm Formation of image by concave mirror : The image distance ( d i ) : 1/ d i = 1/f – 1/ d o = 1/10 – 1/5 = 1/10 – 2/10 = -1/10 d i = -10/1 = -10 cm The minus sign indicates that the image is virtual or the image is behind the mirror. The magnification of image ( m ) : m = – d i / d o = -(-10)/5 = 10/5 = 2 The plus sign indicates that the image is upright. The image height ( h i ) : m = h i / h o h i = h o m = (5 cm)(2) = 10 cm The image height is 10 cm. b) The focal length (f) = 10 cm and the object distance ( d o ) = 15 cm Formation of image by concave mirror : The image distance ( d i ) : 1/ d i = 1/f – 1/ d...

For a concave mirror we know that there are various locations of object and the image being formed. These are as follows. Let F be Focal Point, C its Center of Curvature where C=2F • Object is placed between F and the mirror The image formed is behind the mirror, virtual, upright and Magnified. • Object is placed at the Focal Point Image is formed at infinity. it can be real or virtual upright or inverted depending on approach to F. Formed in front of mirror. • Object is placed between F and C Image is formed between infinity and C. It is real, inverted and magnified. • Object at C Image is formed at C. It is real, inverted and of the same size as the object. • Object beyond center of curvature Image is formed between C and F. It is real, inverted and diminished. • Object at infinity Image is formed at F. t is real, inverted and diminished and tends to be zero. In the problem, it is given that "produces an image whose distance from the mirror is one-third the object distance". Implies that it is under case number 5. The ray diagram for this case is as under Given that #1/3 d_(object)=d_(imag e)# and #f=41cm# Substituting in Mirror and Lens equation #1/d_(object)+1/d_(imag e)=1/f# #1/d_(object)+1/(d_(object)/ 3)=1/41# #implies 1/d_(object)+3/(d_(object))=1/41# #implies 1/d_(object)(1+3)=1/41# #implies d_(object)=4 times41= 164cm# Also #d_(imag e)= 164/3cm#

The correct option is C 15.0 cm and -0.5 Using the lens equation,​ 1/v - 1/u = 1/f​ where, u = 30.0 cm and f = 10.0 cm.​ Solve for v :​ 1/v = 1/f + 1/u ​ 1/v = 1/(10.0 cm) - 1/(30.0 cm) ​ 1/v = 0.0666 c m − 1 v = 1 / (0.0666 c m − 1 ) = 15.0 cm​ ​Then, using the magnification formula,​ m = -v/u to find m​ (u = 30.0 cm; v = 15.0 cm)​ ​Substitute and solve for m:​ m = -(15.0 cm) / (30.0 cm) ​ = - 0.5 Q. The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm, respectively. The distance between the objective and the eyepiece is 15.0 cm. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance, in cm, of the object and the image produced by the objective, measured from the objective lens, are respectively Q. Two thin converging lenses are placed on a common axis, so that the center of one of them coincides with the focus of the other. An object is placed at a distance twice the focal length from the left-hand lens. Where will its image be? What is the lateral magnification? The focal length of each lens is f.

Printable Version Reflection and Mirrors Review Part A: Multiple Choice 1. As the angle of incidence is increased for a ray incident on a reflecting surface, the angle between the incident and reflected rays ultimately approaches what value? a. zero b. 45 degrees c. 90 degrees d. 180 degrees 2. If you stand three feet in front of a plane mirror, how far away would you see yourself in the mirror? a. 1.5 ft b. 3.0 ft c. 6.0 ft d. 12.0 ft 3. A concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis; the corresponding object is located how far from the mirror? a. 20.0 cm b. 15.0 cm c. 7.5 cm d. 5.0 cm 4. A concave mirror forms a real image at 25.0 cm from the mirror surface along the principal axis. If the corresponding object is at a 10.0 cm distance, what is the mirror's focal length? a. 1.4 cm b. 16.7 cm c. 12.4 cm d. 7.1 cm 5. If a virtual image is formed along the principal axis 10.0 cm from a concave mirror with the focal length 15.0 cm, what is the object distance from the mirror? a. 30.0 cm b. 10.0 cm c. 12.4 cm d. 6.0 cm 6. If a virtual image is formed 10.0 cm along the principal axis from a convex mirror of focal length -15.0 cm, what is the object distance from the mirror? a. 30.0 cm b. 10.0 cm c. 6.0 cm d. 3.0 cm 7. If a man's face is 30.0 cm in front of a concave shaving mirror creating an upright image 1.50 times as large as the object, what is the mirror's focal length? a. 12.0 cm b. 20.0 cm c. 70.0 cm d. 90.0 cm 8. ...