An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. draw the ray diagram and find the position, size and the nature of the image formed.

Object distance, u = −25 cm Object height, ho = 5 cm Focal length, f = +10 cm According to the lens formula, `1/v-1/u=1/f` `1/v=1/f+1/u=1/10-1/25=15/250` `v=250/15=16.66 cm` The positive value of v shows that the image is formed at the other side of the lens. Magnification m =`-"image distance"/"object distance"=-v/u=16.66/25=-0.66` The negative sign shows that the image is real and formed behind the lens. Magnification m =`"image height"/"object height"=H_i/H_o=H_i/5` H i=m x H o=-0.66 x 5=-3.3 cm The negative value of image height indicates that the image formed is inverted.The position, size, and nature of image are shown in the following ray diagram. Given: Object distance, u =-25 cm Focal length, f = 10 cm Height of the object, h = 5 cm Applying the lens formula, we get: `1/f=1/v-1/u` ⇒`1/v=1/f+1/u` ⇒`1/v=1/f+1/u` ⇒`11/2/v=1/10+1/-25` ⇒`1/v=(5-2)/50` ⇒`1/v=3/50` ⇒`1/v=3/50` ⇒`v=50/3=+16.6 ` cm The image will be at a distance of 16.6 cm behind the lens. `m=v/u=50/3` `m=h_i/h=50/3` `m=h_i/h=h_i/5` `h_i/5=(-2)/3` h=(-10)/3=-3.3 cm Thus, the image will be 3.3 high. It will be real and inverted

Diagram alongside shows an object AB, placed on the principal axis of lens, such that anerect, enlarged and virtual image is formed. Copy the diagram and answer the following question: (i) Draw an outline of lens. (ii) Draw a straight line from A, which after passing through principal axis is incident on lens and emerges out after reaction. (iii)v Draw another ray, starting from point A and passing through O. (iv) Locate the final image on diagram. Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed : (i) between optical centre and principal focus of a convex lens. (ii) anywhere in front of a concave lens. (iii) at 2F of a convex lens. State the signs and values of magnifications in the above mentioned cases (i) and (ii).

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:An object 5 cm in length is held 25 cm away from a converging lens of focal length10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Here, object size, `h_(1) = 5 cm`, object distance, `u = -25 cm`, focal length of lens, `f = 10 cm` image distance, `v = ?`, image size, `h_(2) = ?` As `(1)/(v) - 1/u = (1)/(v)= (1)/(f), (1)/(f) + 1/u = 1/10 - 1/25 = (5 - 2)/50` or `v = 50/3 = 16.67 cm`. As v is positive, the image formed is real , on the right side of the lens, as shown in Fig. As `m = (h_(2))/(h_(1)) = v/u, (h_(2))/5 = (50//3)/(-25) = (-2)/3` `h_(2) = - 10/ 3 = - 3.3 cm`. Negative sign shows that the image is inverted.

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Here, object size, `h_(1) = 5 cm`, object distance, `u = -25 cm`, focal length of lens, `f = 10 cm` image distance, `v = ?`, image size, `h_(2) = ?` As `(1)/(v) - 1/u = (1)/(v)= (1)/(f), (1)/(f) + 1/u = 1/10 - 1/25 = (5 - 2)/50` or `v = 50/3 = 16.67 cm`. As v is positive, the image formed is real , on the right side of the lens, as shown in Fig. As `m = (h_(2))/(h_(1)) = v/u, (h_(2))/5 = (50//3)/(-25) = (-2)/3` `h_(2) = - 10/ 3 = - 3.3 cm`. Negative sign shows that the image is inverted.

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This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:An object 5 cm in length is held 25 cm away from a converging lens of focal length10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.