<div id=\"tallPI\" class=\"problemIntroduction\">An object is located <span styl

Question

<div id="tallPI" class="problemIntroduction">An object is located <span>26.0</span>&#160;<img src="http://session.masteringphysics.com/render?units=cm" alt="cm" align="middle" /> from a certain lens. The lens forms a real image that is twice as high as the object. This lens has a focal length of 17.3 cm. Now replace the lens. The new lens is a diverging lens whose focal points are at the same distance from the lens as the focal points of the first lens. If the object is <span>5.00</span>&#160;<img src="http://session.masteringphysics.com/render?units=cm" alt="cm" align="middle" /> high, what is the height of the image formed by the new lens? The object is still located <span>26.0</span>&#160;<img src="http://session.masteringphysics.com/render?units=cm" alt="cm" align="middle" /> from the lens.</div>

Explanation / Answer

Distance of the object u = 26.0 cm Magnification m = size of image / size of object = I / O = 2O / O = 2 v / u = 2 from this distance of the image v = 2 u where u = distance of the object from the relation ( 1/ u ) + ( 1/ v ) = ( 1/ f) ( 1/ -u ) + ( 1/ 2u ) = ( 1/ f) ( -1/ 2u ) = ( 1/ f) From this focal length f = -2u = -53 cm

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